Practice · Build values up from the sum rule · Sensim Math

Variant 1 easy

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=3$ and $f(a+b)=f(a)+f(b)+4ab$ . Find the value of $f(4)-f(3)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(4)$ , then subtract.

Givens

$f(1)=3$ .
$f(a+b)=f(a)+f(b)+4ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(4)-f(3)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(4)$ from $2+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+4\times1\times1=3+3+4=10

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+4\times2\times1=10+3+8=21

Each new value is built from ones found earlier.

3Find f(4)

#7 Identify Subproblems 6.EE.A.2

Put

a=2

and

b=2

into the rule.

f(4)=f(2)+f(2)+4\times2\times2=10+10+16=36

4=2+2

uses a value we already have twice.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(4)-f(3)=36-21=15

The function grows, so

f(4)

is bigger than

f(3)

and the difference is positive.

Answer:

15

4 · Reviewdoes it hold up?

Reach $f(4)$ another way: $f(4)=f(3)+f(1)+4\times3\times1=21+3+12=36$ . Same value.

Another way: Look for a pattern: $f(1)=3$ , $f(2)=10$ , $f(3)=21$ , $f(4)=36$ fit $f(n)=2n^2+n$ , and indeed $2(a+b)^2+(a+b)=(2a^2+a)+(2b^2+b)+4ab$ . Then $f(4)-f(3)=36-21=15$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 2 easy

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=2$ and $f(a+b)=f(a)+f(b)+2ab$ . Find the value of $f(3)-f(4)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(4)$ , then subtract.

Givens

$f(1)=2$ .
$f(a+b)=f(a)+f(b)+2ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(3)-f(4)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(4)$ from $2+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+2\times1\times1=2+2+2=6

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+2\times2\times1=6+2+4=12

Each new value is built from ones found earlier.

3Find f(4)

#7 Identify Subproblems 6.EE.A.2

Put

a=2

and

b=2

into the rule.

f(4)=f(2)+f(2)+2\times2\times2=6+6+8=20

4=2+2

uses a value we already have twice.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(3)-f(4)=12-20=-8

The function grows, so

f(4)

is bigger than

f(3)

and the difference is negative.

Answer:

-8

4 · Reviewdoes it hold up?

Reach $f(4)$ another way: $f(4)=f(3)+f(1)+2\times3\times1=12+2+6=20$ . Same value.

Another way: Look for a pattern: $f(1)=2$ , $f(2)=6$ , $f(3)=12$ , $f(4)=20$ fit $f(n)=n^2+n$ , and indeed $(a+b)^2+(a+b)=(a^2+a)+(b^2+b)+2ab$ . Then $f(3)-f(4)=12-20=-8$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 3 easy

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=5$ and $f(a+b)=f(a)+f(b)+2ab$ . Find the value of $f(5)-f(3)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(5)$ , then subtract.

Givens

$f(1)=5$ .
$f(a+b)=f(a)+f(b)+2ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(5)-f(3)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(5)$ from $3+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+2\times1\times1=5+5+2=12

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+2\times2\times1=12+5+4=21

Each new value is built from ones found earlier.

3Find f(5)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=2

into the rule, so we jump straight to

5

.

f(5)=f(3)+f(2)+2\times3\times2=21+12+12=45

5=3+2

uses two values we already have, so there is no need to find

f(4)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(5)-f(3)=45-21=24

The function grows, so

f(5)

is bigger than

f(3)

and the difference is positive.

Answer:

24

4 · Reviewdoes it hold up?

Reach $f(5)$ another way: $f(4)=f(2)+f(2)+2\times2\times2=12+12+8=32$ , and $f(5)=f(4)+f(1)+2\times4\times1=32+5+8=45$ . Same value.

Another way: Look for a pattern: $f(1)=5$ , $f(2)=12$ , $f(3)=21$ , $f(4)=32$ fit $f(n)=n^2+4n$ , and indeed $(a+b)^2+4(a+b)=(a^2+4a)+(b^2+4b)+2ab$ . Then $f(5)-f(3)=45-21=24$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 4 medium

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=4$ and $f(a+b)=f(a)+f(b)+4ab$ . Find the value of $f(3)-f(5)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(5)$ , then subtract.

Givens

$f(1)=4$ .
$f(a+b)=f(a)+f(b)+4ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(3)-f(5)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(5)$ from $3+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+4\times1\times1=4+4+4=12

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+4\times2\times1=12+4+8=24

Each new value is built from ones found earlier.

3Find f(5)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=2

into the rule, so we jump straight to

5

.

f(5)=f(3)+f(2)+4\times3\times2=24+12+24=60

5=3+2

uses two values we already have, so there is no need to find

f(4)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(3)-f(5)=24-60=-36

The function grows, so

f(5)

is bigger than

f(3)

and the difference is negative.

Answer:

-36

4 · Reviewdoes it hold up?

Reach $f(5)$ another way: $f(4)=f(2)+f(2)+4\times2\times2=12+12+16=40$ , and $f(5)=f(4)+f(1)+4\times4\times1=40+4+16=60$ . Same value.

Another way: Look for a pattern: $f(1)=4$ , $f(2)=12$ , $f(3)=24$ , $f(4)=40$ fit $f(n)=2n^2+2n$ , and indeed $2(a+b)^2+2(a+b)=(2a^2+2a)+(2b^2+2b)+4ab$ . Then $f(3)-f(5)=24-60=-36$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 5 easy

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=3$ and $f(a+b)=f(a)+f(b)+2ab$ . Find the value of $f(3)-f(5)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(5)$ , then subtract.

Givens

$f(1)=3$ .
$f(a+b)=f(a)+f(b)+2ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(3)-f(5)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(5)$ from $3+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+2\times1\times1=3+3+2=8

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+2\times2\times1=8+3+4=15

Each new value is built from ones found earlier.

3Find f(5)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=2

into the rule, so we jump straight to

5

.

f(5)=f(3)+f(2)+2\times3\times2=15+8+12=35

5=3+2

uses two values we already have, so there is no need to find

f(4)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(3)-f(5)=15-35=-20

The function grows, so

f(5)

is bigger than

f(3)

and the difference is negative.

Answer:

-20

4 · Reviewdoes it hold up?

Reach $f(5)$ another way: $f(4)=f(2)+f(2)+2\times2\times2=8+8+8=24$ , and $f(5)=f(4)+f(1)+2\times4\times1=24+3+8=35$ . Same value.

Another way: Look for a pattern: $f(1)=3$ , $f(2)=8$ , $f(3)=15$ , $f(4)=24$ fit $f(n)=n^2+2n$ , and indeed $(a+b)^2+2(a+b)=(a^2+2a)+(b^2+2b)+2ab$ . Then $f(3)-f(5)=15-35=-20$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 6 hard

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=6$ and $f(a+b)=f(a)+f(b)+4ab$ . Find the value of $f(3)-f(5)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(5)$ , then subtract.

Givens

$f(1)=6$ .
$f(a+b)=f(a)+f(b)+4ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(3)-f(5)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(5)$ from $3+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+4\times1\times1=6+6+4=16

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+4\times2\times1=16+6+8=30

Each new value is built from ones found earlier.

3Find f(5)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=2

into the rule, so we jump straight to

5

.

f(5)=f(3)+f(2)+4\times3\times2=30+16+24=70

5=3+2

uses two values we already have, so there is no need to find

f(4)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(3)-f(5)=30-70=-40

The function grows, so

f(5)

is bigger than

f(3)

and the difference is negative.

Answer:

-40

4 · Reviewdoes it hold up?

Reach $f(5)$ another way: $f(4)=f(2)+f(2)+4\times2\times2=16+16+16=48$ , and $f(5)=f(4)+f(1)+4\times4\times1=48+6+16=70$ . Same value.

Another way: Look for a pattern: $f(1)=6$ , $f(2)=16$ , $f(3)=30$ , $f(4)=48$ fit $f(n)=2n^2+4n$ , and indeed $2(a+b)^2+4(a+b)=(2a^2+4a)+(2b^2+4b)+4ab$ . Then $f(3)-f(5)=30-70=-40$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 7 medium

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=4$ and $f(a+b)=f(a)+f(b)+2ab$ . Find the value of $f(6)-f(3)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(6)$ , then subtract.

Givens

$f(1)=4$ .
$f(a+b)=f(a)+f(b)+2ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(6)-f(3)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(6)$ from $3+3$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+2\times1\times1=4+4+2=10

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+2\times2\times1=10+4+4=18

Each new value is built from ones found earlier.

3Find f(6)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=3

into the rule, so we jump straight to

6

.

f(6)=f(3)+f(3)+2\times3\times3=18+18+18=54

6=3+3

uses two values we already have, so there is no need to find

f(4)

or

f(5)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(6)-f(3)=54-18=36

The function grows, so

f(6)

is bigger than

f(3)

and the difference is positive.

Answer:

36

4 · Reviewdoes it hold up?

Reach $f(6)$ another way: $f(4)=f(2)+f(2)+2\times2\times2=10+10+8=28$ , and $f(6)=f(4)+f(2)+2\times4\times2=28+10+16=54$ . Same value.

Another way: Look for a pattern: $f(1)=4$ , $f(2)=10$ , $f(3)=18$ , $f(4)=28$ fit $f(n)=n^2+3n$ , and indeed $(a+b)^2+3(a+b)=(a^2+3a)+(b^2+3b)+2ab$ . Then $f(6)-f(3)=54-18=36$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 8 hard

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=5$ and $f(a+b)=f(a)+f(b)+6ab$ . Find the value of $f(3)-f(4)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(4)$ , then subtract.

Givens

$f(1)=5$ .
$f(a+b)=f(a)+f(b)+6ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(3)-f(4)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(4)$ from $2+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+6\times1\times1=5+5+6=16

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+6\times2\times1=16+5+12=33

Each new value is built from ones found earlier.

3Find f(4)

#7 Identify Subproblems 6.EE.A.2

Put

a=2

and

b=2

into the rule.

f(4)=f(2)+f(2)+6\times2\times2=16+16+24=56

4=2+2

uses a value we already have twice.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(3)-f(4)=33-56=-23

The function grows, so

f(4)

is bigger than

f(3)

and the difference is negative.

Answer:

-23

4 · Reviewdoes it hold up?

Reach $f(4)$ another way: $f(4)=f(3)+f(1)+6\times3\times1=33+5+18=56$ . Same value.

Another way: Look for a pattern: $f(1)=5$ , $f(2)=16$ , $f(3)=33$ , $f(4)=56$ fit $f(n)=3n^2+2n$ , and indeed $3(a+b)^2+2(a+b)=(3a^2+2a)+(3b^2+2b)+6ab$ . Then $f(3)-f(4)=33-56=-23$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 9 hard

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=6$ and $f(a+b)=f(a)+f(b)+2ab$ . Find the value of $f(4)-f(3)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(4)$ , then subtract.

Givens

$f(1)=6$ .
$f(a+b)=f(a)+f(b)+2ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(4)-f(3)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(4)$ from $2+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+2\times1\times1=6+6+2=14

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+2\times2\times1=14+6+4=24

Each new value is built from ones found earlier.

3Find f(4)

#7 Identify Subproblems 6.EE.A.2

Put

a=2

and

b=2

into the rule.

f(4)=f(2)+f(2)+2\times2\times2=14+14+8=36

4=2+2

uses a value we already have twice.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(4)-f(3)=36-24=12

The function grows, so

f(4)

is bigger than

f(3)

and the difference is positive.

Answer:

12

4 · Reviewdoes it hold up?

Reach $f(4)$ another way: $f(4)=f(3)+f(1)+2\times3\times1=24+6+6=36$ . Same value.

Another way: Look for a pattern: $f(1)=6$ , $f(2)=14$ , $f(3)=24$ , $f(4)=36$ fit $f(n)=n^2+5n$ , and indeed $(a+b)^2+5(a+b)=(a^2+5a)+(b^2+5b)+2ab$ . Then $f(4)-f(3)=36-24=12$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 10 medium

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=2$ and $f(a+b)=f(a)+f(b)+2ab$ . Find the value of $f(3)-f(6)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(6)$ , then subtract.

Givens

$f(1)=2$ .
$f(a+b)=f(a)+f(b)+2ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(3)-f(6)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(6)$ from $3+3$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+2\times1\times1=2+2+2=6

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+2\times2\times1=6+2+4=12

Each new value is built from ones found earlier.

3Find f(6)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=3

into the rule, so we jump straight to

6

.

f(6)=f(3)+f(3)+2\times3\times3=12+12+18=42

6=3+3

uses two values we already have, so there is no need to find

f(4)

or

f(5)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(3)-f(6)=12-42=-30

The function grows, so

f(6)

is bigger than

f(3)

and the difference is negative.

Answer:

-30

4 · Reviewdoes it hold up?

Reach $f(6)$ another way: $f(4)=f(2)+f(2)+2\times2\times2=6+6+8=20$ , and $f(6)=f(4)+f(2)+2\times4\times2=20+6+16=42$ . Same value.

Another way: Look for a pattern: $f(1)=2$ , $f(2)=6$ , $f(3)=12$ , $f(4)=20$ fit $f(n)=n^2+n$ , and indeed $(a+b)^2+(a+b)=(a^2+a)+(b^2+b)+2ab$ . Then $f(3)-f(6)=12-42=-30$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 11 medium

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=5$ and $f(a+b)=f(a)+f(b)+4ab$ . Find the value of $f(3)-f(6)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(6)$ , then subtract.

Givens

$f(1)=5$ .
$f(a+b)=f(a)+f(b)+4ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(3)-f(6)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(6)$ from $3+3$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+4\times1\times1=5+5+4=14

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+4\times2\times1=14+5+8=27

Each new value is built from ones found earlier.

3Find f(6)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=3

into the rule, so we jump straight to

6

.

f(6)=f(3)+f(3)+4\times3\times3=27+27+36=90

6=3+3

uses two values we already have, so there is no need to find

f(4)

or

f(5)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(3)-f(6)=27-90=-63

The function grows, so

f(6)

is bigger than

f(3)

and the difference is negative.

Answer:

-63

4 · Reviewdoes it hold up?

Reach $f(6)$ another way: $f(4)=f(2)+f(2)+4\times2\times2=14+14+16=44$ , and $f(6)=f(4)+f(2)+4\times4\times2=44+14+32=90$ . Same value.

Another way: Look for a pattern: $f(1)=5$ , $f(2)=14$ , $f(3)=27$ , $f(4)=44$ fit $f(n)=2n^2+3n$ , and indeed $2(a+b)^2+3(a+b)=(2a^2+3a)+(2b^2+3b)+4ab$ . Then $f(3)-f(6)=27-90=-63$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Variant 12 hard

The variables $x$ and $y$ are both natural numbers, and the function $y=f(x)$ satisfies the two conditions $f(1)=4$ and $f(a+b)=f(a)+f(b)+6ab$ . Find the value of $f(5)-f(3)$ .

Your answer

Show solution

1 · Understandwhat's really being asked

We only know $f(1)$ and a rule that gives $f(a+b)$ from $f(a)$ and $f(b)$ . Use the rule to reach $f(3)$ and $f(5)$ , then subtract.

Givens

$f(1)=4$ .
$f(a+b)=f(a)+f(b)+6ab$ for natural numbers $a$ and $b$ .

Unknowns

The value of $f(5)-f(3)$ .

Constraints

Inputs and outputs are natural numbers, so we can only plug in $a, b = 1, 2, 3, \dots$

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#5 Look for a Pattern

The rule turns any input into a sum of smaller inputs. Split the job into small pieces: get $f(2)$ from $1+1$ , then $f(3)$ from $2+1$ , then $f(5)$ from $3+2$ , each time substituting known values into the rule.

3 · Execute4 carry out the plan

1Find f(2)

#13 Convert to Algebra 8.F.A.1

Put

a=1

and

b=1

into the rule.

f(2)=f(1)+f(1)+6\times1\times1=4+4+6=14

2=1+1

, and both pieces are values we already know.

2Find f(3)

#13 Convert to Algebra 6.EE.A.2

Put

a=2

and

b=1

into the rule.

f(3)=f(2)+f(1)+6\times2\times1=14+4+12=30

Each new value is built from ones found earlier.

3Find f(5)

#7 Identify Subproblems 6.EE.A.2

Put

a=3

and

b=2

into the rule, so we jump straight to

5

.

f(5)=f(3)+f(2)+6\times3\times2=30+14+36=80

5=3+2

uses two values we already have, so there is no need to find

f(4)

.

4Subtract

#7 Identify Subproblems 6.EE.A.2

Now compute the difference.

f(5)-f(3)=80-30=50

The function grows, so

f(5)

is bigger than

f(3)

and the difference is positive.

Answer:

50

4 · Reviewdoes it hold up?

Reach $f(5)$ another way: $f(4)=f(2)+f(2)+6\times2\times2=14+14+24=52$ , and $f(5)=f(4)+f(1)+6\times4\times1=52+4+24=80$ . Same value.

Another way: Look for a pattern: $f(1)=4$ , $f(2)=14$ , $f(3)=30$ , $f(4)=52$ fit $f(n)=3n^2+n$ , and indeed $3(a+b)^2+(a+b)=(3a^2+a)+(3b^2+b)+6ab$ . Then $f(5)-f(3)=80-30=50$ .

Standardsmin grade 8

8.F.A.1 Understand that a function is a rule that assigns exactly one output to each input — Reading the rule as a function that gives exactly one output for each natural-number input.
6.EE.A.2 Write, read, and evaluate expressions with variables — Substituting values for $a$ and $b$ and evaluating the expression.

💡Takeaway. When a rule builds big inputs out of small ones, climb up from the value you know.

Build values up from the sum rule · 12 practice problems

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