Mathematical-Reasoning · Unit 12 · Video 3 · Interactive Practice

Induction on Sums: Add the Next Term to Both Sides

IKey Formulas

Statement	Formula
Sum of the first $n$ squares	$\displaystyle\sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6}$
Sum of the first $n$ natural numbers	$\displaystyle\sum_{j=1}^{n} j = \frac{n(n+1)}{2}$
The link: add the last term	$\displaystyle\sum_{j=1}^{k+1} j^2 = \left(\sum_{j=1}^{k} j^2\right) + (k+1)^2$
Goal $P(k+1)$ , brackets simplified	$\displaystyle\sum_{j=1}^{k+1} j^2 = \frac{(k+1)(k+2)(2k+3)}{6}$

Key Insight: Compare the left sides of $P(k)$ and $P(k+1)$ to find the extra term, write the goal $P(k+1)$ with its right side simplified, add the extra term to both sides of the inductive hypothesis, then rewrite the right side until it matches the goal. The proof opens by announcing induction and defining $P(n)$ , so the reader knows what each step must establish.

IIOne More Term on Both Sides

The left side of $P(k+1)$ is the left side of $P(k)$ plus one more term.

Sum k 2

IIIRewrite the Right Side Until It Matches the Goal

Each rewrite of the right side equals the one before it, so at any $k$ every rewritten row has the same value.

Proof Step 0 k 3

(1) inductive hypothesis P(k)

1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}

(2) goal P(k+1), brackets simplified

1^2 + 2^2 + \cdots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}

step 1 · add (k+1)² to both sides of (1)

1^2 + \cdots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2

step 2 · put both terms over 6

= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}

step 3 · factor out k+1

= \frac{(k+1)\big[k(2k+1) + 6(k+1)\big]}{6}

step 4 · expand the bracket: 2k² + k + 6k + 6

= \frac{(k+1)(2k^2 + 7k + 6)}{6}

step 5 · factor the quadratic: the right side of (2)

= \frac{(k+1)(k+2)(2k+3)}{6}

(3) inductive hypothesis P(k)

1 + 2 + \cdots + k = \frac{k(k+1)}{2}

(4) goal P(k+1)

1 + 2 + \cdots + k + (k+1) = \frac{(k+1)(k+2)}{2}

step 1 · add k+1 to both sides of (3)

1 + 2 + \cdots + k + (k+1) = \frac{k(k+1)}{2} + (k+1)

step 2 · common denominator 2

= \frac{k(k+1) + 2(k+1)}{2}

step 3 · expand the numerator: k² + k + 2k + 2

= \frac{k^2 + 3k + 2}{2}

step 4 · factor: the right side of (4)

= \frac{(k+1)(k+2)}{2}

💡 The numbers check one $k$ at a time, which is evidence; the algebra is the proof, because each rewrite holds for every natural number $k$ .

IVQuiz Questions

Problem 1 · Find the Extra Term

Given: for each natural number $n$ , let $P(n)$ be $1 + 3 + 5 + \cdots + (2n-1) = n^2$ , the sum of the first $n$ odd natural numbers. Which term must be added to both sides of $P(k)$ so that its left side becomes the left side of $P(k+1)$ ?

✅ Correct! The left side of

P(k+1)

ends with the term for

n = k+1

, which is

2(k+1) - 1 = 2k + 1

❌ Not quite.

2k - 1

is the last term of

P(k)

itself, already on its left side. The next term replaces

k

k + 1

2k - 1

❌ Not quite.

k + 1

is the next term of

1 + 2 + \cdots + k

. Here the terms are odd numbers

2j - 1

, so the next one is

2(k+1) - 1

❌ Not quite.

2(k+1) - 1 = 2k + 1

, not

2k + 3

. The term

2k + 3

is the one for

n = k + 2

, one step too far.

Show solution

Write out the two left sides, with the term the dots hide in $P(k+1)$ shown:

P(k):\quad 1 + 3 + \cdots + (2k-1)

P(k+1):\quad 1 + 3 + \cdots + (2k-1) + \big(2(k+1) - 1\big)

The first part of the left side of $P(k+1)$ is exactly the left side of $P(k)$ . The extra term is $2(k+1) - 1 = 2k + 1$ . Adding it to both sides of $P(k)$ gives

1 + 3 + \cdots + (2k-1) + (2k+1) = k^2 + 2k + 1 = (k+1)^2,

which is $P(k+1)$ .

Problem 2 · Write the Goal

Given: for each natural number $n$ , let $P(n)$ be $1^3 + 2^3 + \cdots + n^3 = \dfrac{n^2(n+1)^2}{4}$ . In the inductive step, what is the right side of the goal $P(k+1)$ , with its brackets simplified?

✅ Correct! Replace every

n

k+1

n^2

becomes

(k+1)^2

and

(n+1)^2

becomes

\big((k+1)+1\big)^2 = (k+2)^2

❌ Not quite.

P(k+1)

is not

P(k)

with

1

added. It is

P(n)

with every

n

replaced by

k + 1

, on both sides.

❌ Not quite. The goal is the formula with every

n

replaced by

k+1

. Adding a term to the hypothesis is how the proof reaches it, and the term to add is

(k+1)^3

, not

k + 1

❌ Close. Only the first

n

was replaced. The factor

(n+1)^2

must become

\big((k+1)+1\big)^2 = (k+2)^2

too.

Show solution

$P(k+1)$ is $P(n)$ with $n$ replaced by $k+1$ everywhere:

1^3 + 2^3 + \cdots + (k+1)^3 = \frac{(k+1)^2\big[(k+1)+1\big]^2}{4}.

Simplifying the bracket, $(k+1) + 1 = k + 2$ , so the goal is

1^3 + 2^3 + \cdots + k^3 + (k+1)^3 = \frac{(k+1)^2(k+2)^2}{4}.

Check at $k = 1$ : the left side is $1 + 8 = 9$ and the right side is $\frac{4 \cdot 9}{4} = 9$ .

Problem 3 · Run the Inductive Step

Given: for each natural number $n$ , let $P(n)$ be $1 \cdot 2 + 2 \cdot 3 + \cdots + n(n+1) = \dfrac{n(n+1)(n+2)}{3}$ . Assume $P(k)$ for a natural number $k$ .

Which term is added to both sides of P(k)?

After adding it, what does the right side simplify to?

✅ Correct! The right side becomes

\frac{(k+1)(k+2)(k+3)}{3}

, which is the right side of

P(k+1)

, so

P(k)

implies

P(k+1)

❌ Not quite.

k(k+1)

is the last term of

P(k)

, already on its left side. The next term is

n(n+1)

n = k+1

❌ Not quite. Replace both factors:

n(n+1)

n = k+1

(k+1)\big((k+1)+1\big) = (k+1)(k+2)

❌ Not quite.

(k+2)(k+3)

is the term for

n = k+2

. The term for

n = k+1

(k+1)(k+2)

❌ Not quite. Find the term of the sum at

n = k+1

❌ Not quite. Over the common denominator

3

, the added term becomes

\frac{3(k+1)(k+2)}{3}

; this option adds

(k+1)(k+2)

to the numerator without the factor

3

❌ Not quite. The denominator stays

3

\frac{k(k+1)(k+2) + 3(k+1)(k+2)}{3}

. The

6

belongs to the sum of squares formula.

❌ Not quite.

2k+3

comes from the sum of squares proof. Here factoring

(k+1)(k+2)

out of

k(k+1)(k+2) + 3(k+1)(k+2)

leaves

k + 3

❌ Not quite. Put both terms over

3

and factor out

(k+1)(k+2)

Show solution

The left side of $P(k+1)$ is the left side of $P(k)$ plus the term for $n = k+1$ , which is $(k+1)(k+2)$ . The goal $P(k+1)$ , with brackets simplified, is

1 \cdot 2 + \cdots + k(k+1) + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}.

Add $(k+1)(k+2)$ to both sides of $P(k)$ . The left side becomes the left side of the goal, and the right side is

\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{k(k+1)(k+2) + 3(k+1)(k+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}.

That is the right side of the goal, so if $P(k)$ is true, then $P(k+1)$ is true. Check at $k = 2$ : $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 = 20$ and $\frac{3 \cdot 4 \cdot 5}{3} = 20$ .

Problem 4 · The Link in Summation Notation

Given: for each natural number $n$ , let $P(n)$ be $\displaystyle\sum_{j=1}^{n} (3j - 2) = \frac{n(3n-1)}{2}$ .

Complete the link: the sum from j = 1 to k + 1 equals the sum from j = 1 to k, plus …

Adding it to both sides of P(k) and writing the right side over 2, what is the numerator?

✅ Correct!

3k^2 + 5k + 2 = (k+1)(3k+2)

, and

3(k+1) - 1 = 3k + 2

, so the right side is

\frac{(k+1)\big(3(k+1)-1\big)}{2}

, the right side of

P(k+1)

❌ Not quite.

3k - 2

is the term for

j = k

, the last term of the sum up to

k

. The sum up to

k+1

adds the term for

j = k+1

❌ Not quite. The index

j

appears only inside the sum. The added term is a number depending on

k

: the term

3j - 2

j = k+1

❌ Not quite.

3k + 3

3(k+1)

; the

-2

is missing. The term is

3(k+1) - 2

❌ Not quite. Evaluate

3j - 2

j = k+1

❌ Not quite. Over the denominator

2

, the added term is

\frac{2(3k+1)}{2}

; this numerator adds

3k + 1

without the factor

2

❌ Close.

2(3k+1) = 6k + 2

; only the

3k

was doubled. The numerator is

3k^2 - k + 6k + 2

❌ Not quite. Check the sign:

k(3k - 1) = 3k^2 - k

, so the numerator is

3k^2 - k + 6k + 2

❌ Not quite. Expand

k(3k-1) + 2(3k+1)

Show solution

The term of the sum at $j = k+1$ is $3(k+1) - 2 = 3k + 1$ , so for each natural number $k$ ,

\sum_{j=1}^{k+1} (3j - 2) = \left(\sum_{j=1}^{k} (3j - 2)\right) + (3k + 1).

The goal $P(k+1)$ , with its bracket simplified ( $3(k+1) - 1 = 3k + 2$ ), is $\displaystyle\sum_{j=1}^{k+1} (3j - 2) = \frac{(k+1)(3k+2)}{2}$ .

Add $3k + 1$ to both sides of $P(k)$ . The right side becomes

\frac{k(3k-1)}{2} + (3k+1) = \frac{3k^2 - k + 6k + 2}{2} = \frac{3k^2 + 5k + 2}{2} = \frac{(k+1)(3k+2)}{2}.

Check the factoring: $(k+1)(3k+2) = 3k^2 + 2k + 3k + 2$ . That is the right side of the goal, so $P(k)$ implies $P(k+1)$ .

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