Practice · Pattern in equations predicts the answer · Sensim Math

Variant 1 easy

Find the rule in the calculations, then write the fifth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the fifth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The fifth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the fifth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the fifth calculation has 9 terms.

1,\ 3,\ 5,\ 7,\ 9 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The fifth calculation has 9 terms, so it runs through the first 9 odd numbers, ending at the 9th odd number, which is 2 x 9 - 1 = 17.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The fifth calculation has 9 terms, so its result is 9 x 9.

9 \times 9 = 81

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17 = 81

4 · Reviewdoes it hold up?

The result must be a square: 81 = 9 x 9, and the calculation uses 9 terms ending at 17, both matching the pattern. It is larger than the fourth calculation's 49, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 81.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the fifth calculation has 9 terms ending at 17.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 9 x 9 = 81.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the fifth calculation (9 odd numbers) equals 9 x 9 = 81 with no long adding!

Variant 2 easy

Find the rule in the calculations, then write the sixth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the sixth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The sixth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the sixth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the sixth calculation has 11 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The sixth calculation has 11 terms, so it runs through the first 11 odd numbers, ending at the 11th odd number, which is 2 x 11 - 1 = 21.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The sixth calculation has 11 terms, so its result is 11 x 11.

11 \times 11 = 121

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21 = 121

4 · Reviewdoes it hold up?

The result must be a square: 121 = 11 x 11, and the calculation uses 11 terms ending at 21, both matching the pattern. It is larger than the fourth calculation's 49, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 121.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the sixth calculation has 11 terms ending at 21.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 11 x 11 = 121.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the sixth calculation (11 odd numbers) equals 11 x 11 = 121 with no long adding!

Variant 3 easy

Find the rule in the calculations, then write the seventh calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the seventh calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The seventh calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the seventh calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the seventh calculation has 13 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The seventh calculation has 13 terms, so it runs through the first 13 odd numbers, ending at the 13th odd number, which is 2 x 13 - 1 = 25.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The seventh calculation has 13 terms, so its result is 13 x 13.

13 \times 13 = 169

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25 = 169

4 · Reviewdoes it hold up?

The result must be a square: 169 = 13 x 13, and the calculation uses 13 terms ending at 25, both matching the pattern. It is larger than the fourth calculation's 49, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 169.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the seventh calculation has 13 terms ending at 25.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 13 x 13 = 169.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the seventh calculation (13 odd numbers) equals 13 x 13 = 169 with no long adding!

Variant 4 hard

Find the rule in the calculations, then write the twelfth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the twelfth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The twelfth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the twelfth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the twelfth calculation has 23 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15,\ 17,\ 19,\ 21,\ 23 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The twelfth calculation has 23 terms, so it runs through the first 23 odd numbers, ending at the 23th odd number, which is 2 x 23 - 1 = 45.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The twelfth calculation has 23 terms, so its result is 23 x 23.

23 \times 23 = 529

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37+39+41+43+45 = 529

4 · Reviewdoes it hold up?

The result must be a square: 529 = 23 x 23, and the calculation uses 23 terms ending at 45, both matching the pattern. It is larger than the fifth calculation's 81, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 529.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the twelfth calculation has 23 terms ending at 45.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 23 x 23 = 529.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the twelfth calculation (23 odd numbers) equals 23 x 23 = 529 with no long adding!

Variant 5 medium

Find the rule in the calculations, then write the eighth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the eighth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The eighth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the eighth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the eighth calculation has 15 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The eighth calculation has 15 terms, so it runs through the first 15 odd numbers, ending at the 15th odd number, which is 2 x 15 - 1 = 29.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The eighth calculation has 15 terms, so its result is 15 x 15.

15 \times 15 = 225

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29 = 225

4 · Reviewdoes it hold up?

The result must be a square: 225 = 15 x 15, and the calculation uses 15 terms ending at 29, both matching the pattern. It is larger than the fifth calculation's 81, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 225.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the eighth calculation has 15 terms ending at 29.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 15 x 15 = 225.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the eighth calculation (15 odd numbers) equals 15 x 15 = 225 with no long adding!

Variant 6 medium

Find the rule in the calculations, then write the seventh calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the seventh calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The seventh calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the seventh calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the seventh calculation has 13 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The seventh calculation has 13 terms, so it runs through the first 13 odd numbers, ending at the 13th odd number, which is 2 x 13 - 1 = 25.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The seventh calculation has 13 terms, so its result is 13 x 13.

13 \times 13 = 169

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25 = 169

4 · Reviewdoes it hold up?

The result must be a square: 169 = 13 x 13, and the calculation uses 13 terms ending at 25, both matching the pattern. It is larger than the fifth calculation's 81, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 169.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the seventh calculation has 13 terms ending at 25.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 13 x 13 = 169.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the seventh calculation (13 odd numbers) equals 13 x 13 = 169 with no long adding!

Variant 7 medium

Find the rule in the calculations, then write the sixth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the sixth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The sixth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the sixth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the sixth calculation has 11 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The sixth calculation has 11 terms, so it runs through the first 11 odd numbers, ending at the 11th odd number, which is 2 x 11 - 1 = 21.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The sixth calculation has 11 terms, so its result is 11 x 11.

11 \times 11 = 121

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21 = 121

4 · Reviewdoes it hold up?

The result must be a square: 121 = 11 x 11, and the calculation uses 11 terms ending at 21, both matching the pattern. It is larger than the fifth calculation's 81, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 121.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the sixth calculation has 11 terms ending at 21.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 11 x 11 = 121.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the sixth calculation (11 odd numbers) equals 11 x 11 = 121 with no long adding!

Variant 8 medium

Find the rule in the calculations, then write the tenth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the tenth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The tenth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the tenth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the tenth calculation has 19 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15,\ 17,\ 19 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The tenth calculation has 19 terms, so it runs through the first 19 odd numbers, ending at the 19th odd number, which is 2 x 19 - 1 = 37.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The tenth calculation has 19 terms, so its result is 19 x 19.

19 \times 19 = 361

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37 = 361

4 · Reviewdoes it hold up?

The result must be a square: 361 = 19 x 19, and the calculation uses 19 terms ending at 37, both matching the pattern. It is larger than the fifth calculation's 81, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 361.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the tenth calculation has 19 terms ending at 37.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 19 x 19 = 361.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the tenth calculation (19 odd numbers) equals 19 x 19 = 361 with no long adding!

Variant 9 hard

Find the rule in the calculations, then write the eleventh calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the eleventh calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The eleventh calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the eleventh calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the eleventh calculation has 21 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15,\ 17,\ 19,\ 21 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The eleventh calculation has 21 terms, so it runs through the first 21 odd numbers, ending at the 21th odd number, which is 2 x 21 - 1 = 41.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The eleventh calculation has 21 terms, so its result is 21 x 21.

21 \times 21 = 441

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37+39+41 = 441

4 · Reviewdoes it hold up?

The result must be a square: 441 = 21 x 21, and the calculation uses 21 terms ending at 41, both matching the pattern. It is larger than the fifth calculation's 81, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 441.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the eleventh calculation has 21 terms ending at 41.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 21 x 21 = 441.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the eleventh calculation (21 odd numbers) equals 21 x 21 = 441 with no long adding!

Variant 10 medium

Find the rule in the calculations, then write the ninth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the ninth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The ninth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the ninth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the ninth calculation has 17 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15,\ 17 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The ninth calculation has 17 terms, so it runs through the first 17 odd numbers, ending at the 17th odd number, which is 2 x 17 - 1 = 33.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The ninth calculation has 17 terms, so its result is 17 x 17.

17 \times 17 = 289

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33 = 289

4 · Reviewdoes it hold up?

The result must be a square: 289 = 17 x 17, and the calculation uses 17 terms ending at 33, both matching the pattern. It is larger than the fifth calculation's 81, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 289.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the ninth calculation has 17 terms ending at 33.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 17 x 17 = 289.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the ninth calculation (17 odd numbers) equals 17 x 17 = 289 with no long adding!

Variant 11 hard

Find the rule in the calculations, then write the eighth calculation.

Position	Calculation
1st	$1$
2nd	$1+3+5 = 9$
3rd	$1+3+5+7+9 = 25$
4th	$1+3+5+7+9+11+13 = 49$
5th	$1+3+5+7+9+11+13+15+17 = 81$
6th	$1+3+5+7+9+11+13+15+17+19+21 = 121$

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

A list of calculations adds up consecutive odd numbers starting from 1. The 1st calculation is just 1, the 2nd is 1+3+5=9, the 3rd is 1+3+5+7+9=25, and each calculation uses more odd numbers than the one before. We must write the eighth calculation, including its result, without grinding through every sum.

Givens

1st: 1
2nd: 1+3+5 = 9
3rd: 1+3+5+7+9 = 25
Each later calculation adds two more odd numbers than the previous one.

Unknowns

The eighth calculation: which odd numbers it adds and what it equals.

Constraints

Each calculation adds consecutive odd numbers starting at 1.
The number of terms grows by 2 each step: 1, 3, 5, 7, 9, ...

2 · Planchoose the strategy

#5 Look for a Pattern · also uses: #9 Solve an Easier Related Problem

The results 1, 9, 25, 49, 81 are the square numbers 1, 3, 5, 7, 9 squared, and the number of terms grows by 2 each step. Spotting these two patterns lets us write the eighth calculation and its result directly.

3 · Execute3 carry out the plan

1Count the terms in each calculation

#5 Look for a Pattern 4.OA.C.5

Count how many odd numbers each calculation adds: the term count goes up by 2 each time (1, 3, 5, ...), so the eighth calculation has 15 terms.

1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15 \text{ terms}

Each step just tacks on the next two odd numbers, so the count of terms climbs by 2 every time.

2Find the last odd number used

#5 Look for a Pattern 4.OA.C.5

The eighth calculation has 15 terms, so it runs through the first 15 odd numbers, ending at the 15th odd number, which is 2 x 15 - 1 = 29.

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 + 29

The odd numbers in order are 1, 3, 5, ...; that fixes where the sum stops.

3Use the square-number pattern for the result

#9 Solve an Easier Related Problem 3.OA.D.9

Look at the results: 1 = 1x1, 9 = 3x3, 25 = 5x5. The result is always the number of terms multiplied by itself. The eighth calculation has 15 terms, so its result is 15 x 15.

15 \times 15 = 225

Adding the first few odd numbers always builds a perfect square, so we can predict the total without adding them all.

Answer:

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29 = 225

4 · Reviewdoes it hold up?

The result must be a square: 225 = 15 x 15, and the calculation uses 15 terms ending at 29, both matching the pattern. It is larger than the sixth calculation's 121, which is right since we added more odd numbers.

Another way: Pair the terms from the outside in: each outer pair sums to the same total, and pairing all the way to the middle gives the same square 225.

Standardsmin grade 4

4.OA.C.5 Generate a number or shape pattern following a given rule — Extending the term-count pattern to find the eighth calculation has 15 terms ending at 29.
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations — Recognizing the results as squares to predict 15 x 15 = 225.

💡Takeaway. Adding odd numbers starting at 1 always makes a square, so the eighth calculation (15 odd numbers) equals 15 x 15 = 225 with no long adding!

Pattern in equations predicts the answer · 11 practice problems

Generated variants — 11

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