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Module 5 · L6 of 20 ~45 min ⚡ +95 XP available

Induction for Sum of Squares

A staircase of square tiles: 1 on the first step, 4 on the second, 9 on the third. How many tiles in total after $n$ steps? The formula $\dfrac{n(n+1)(2n+1)}{6}$ answers this instantly — but proving it by induction is more algebraically demanding than any sum proof you've seen so far. Mastering the factorisation step here will prepare you for every polynomial induction proof to come.

Today's hook — Without any formula, estimate: what is $1^2 + 2^2 + 3^2 + 4^2 + 5^2$? Just add them up mentally. Jot your answer. You'll verify it with the formula after card 05.

0/5QUESTS

Recall — your gut answer first

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Calculate $1^2 + 2^2 + 3^2 + 4^2 + 5^2$ by direct addition (no formula). Then estimate what you think $1^2 + 2^2 + \cdots + 10^2$ equals. Write both below.

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The big idea

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This proof is more algebraically demanding than simpler series because the inductive step requires factorising a quadratic. The two key moves are: factor out $(k+1)$ early, then factorise $2k^2 + 7k + 6$ into $(k+2)(2k+3)$.

The sum of squares formula is:

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

In the $k+1$ step: factor out $(k+1)$, collect $k(2k+1) + 6(k+1) = 2k^2+7k+6 = (k+2)(2k+3)$.

$\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

Factor $(k+1)$ first

Take out $(k+1)$ as a common factor before combining terms. This keeps the algebra manageable and reveals the quadratic clearly.

Check the factorisation

After factorising $2k^2+7k+6 = (k+2)(2k+3)$, expand it back: $(k+2)(2k+3) = 2k^2+3k+4k+6 = 2k^2+7k+6$. Always verify.

Check the target

Write the target for $k+1$ before starting: $\dfrac{(k+1)(k+2)(2k+3)}{6}$. Your final line must match this exactly.

What you'll master

Know

Key facts

$1^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all $n \geq 1$
$(k+1)^2 = k^2 + 2k + 1$ (not $k^2 + 1$)
$2k^2 + 7k + 6$ factorises as $(k+2)(2k+3)$

Understand

Concepts

Why factoring out $(k+1)$ before expanding simplifies the algebra
How to identify the correct common denominator (6) when combining fractions
Why you must verify the factorisation by expanding back out

Can do

Skills

Write a complete induction proof for the sum of squares formula
Evaluate $1^2 + 2^2 + \cdots + n^2$ using the formula for any given $n$
Avoid the three most common algebraic errors in this proof

Key terms

Sum of squaresThe series $1^2 + 2^2 + 3^2 + \cdots + n^2$, where the $k$-th term is $k^2$.

Common denominatorIn this proof, the denominator is 6 throughout. Multiply $(k+1)^2$ by $\frac{6}{6}$ to combine with the hypothesis fraction.

$(k+1)$ as common factorBoth $k(k+1)(2k+1)/6$ and $(k+1)^2$ contain $(k+1)$. Factor it out to simplify before expanding.

Quadratic factorisationWriting $2k^2+7k+6 = (k+2)(2k+3)$. Verified by expanding the right side.

Target expressionThe RHS of the $k+1$ case: $\dfrac{(k+1)(k+2)(2k+3)}{6}$. Write this before starting Step 3.

Inductive conclusionThe final sentence of the proof, stating that the result holds for all $n \geq 1$ by the principle of mathematical induction.

The sum of squares formula

core concept

The sum of the squares of the first $n$ positive integers is:

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

This formula appears in statistics (sum of squared deviations), physics (moments of inertia), and many HSC exam questions. Unlike the geometric or arithmetic series proofs, the inductive step here requires factorising a quadratic — the key new skill.

Verify your hook: For $n = 5$:

$\dfrac{5 \times 6 \times 11}{6} = \dfrac{330}{6} = \mathbf{55}$

So $1 + 4 + 9 + 16 + 25 = 55$. Does this match your direct addition?

Quick evaluation trick. The formula $\dfrac{n(n+1)(2n+1)}{6}$ has three factors in the numerator. Note that one of $n$, $n+1$, $n+2$ is always divisible by 2 and one of $n$, $n+1$, $n+2$ is always divisible by 3, so the result is always an integer — a useful sanity check.

Sum of squares: $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$

Pause — copy the sum-of-squares formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ and verify it for $n=1$ and $n=2$ into your book.

Quick check: Which expression gives $1^2 + 2^2 + 3^2 + \cdots + 10^2$?

The complete induction proof

core concept

We just saw that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$. That raises a question: how do you simplify $\frac{k(k+1)(2k+1)}{6}+(k+1)^2$ to reach $\frac{(k+1)(k+2)(2k+3)}{6}$? This card answers it → by factorising $(k+1)$ from the numerator and verifying $k(2k+1)+6(k+1)=(k+2)(2k+3)$.

We prove that $1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all positive integers $n$.

Step 1 — Base case ($n = 1$):

LHS $= 1^2 = 1$

RHS $= \dfrac{1 \times 2 \times 3}{6} = \dfrac{6}{6} = 1$

LHS $=$ RHS. True for $n = 1$.

Step 2 — Inductive hypothesis:

Assume the statement is true for $n = k$ (where $k \geq 1$):

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

Step 3 — Inductive step ($n = k + 1$):

Target (write this first): $\dfrac{(k+1)(k+2)(2k+3)}{6}$

LHS$_{k+1} = \underbrace{\dfrac{k(k+1)(2k+1)}{6}}_{\text{hypothesis}} + (k+1)^2$

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

$= \dfrac{(k+1)\bigl[k(2k+1) + 6(k+1)\bigr]}{6}$ (factor out $(k+1)$)

$= \dfrac{(k+1)\bigl[2k^2 + k + 6k + 6\bigr]}{6}$

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

Factorise: $2k^2 + 7k + 6 = (k+2)(2k+3)$

$= \dfrac{(k+1)(k+2)(2k+3)}{6}$ $=$ RHS$_{k+1}$. $\checkmark$

By the principle of mathematical induction, the statement is true for all $n \geq 1$. $\blacksquare$

Factorisation check. $(k+2)(2k+3) = 2k^2 + 3k + 4k + 6 = 2k^2 + 7k + 6$. Always verify your factorisation before writing $\checkmark$ on the inductive step.

We prove that $1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all positive integers $n$.

Pause — copy the induction proof, highlighting the factorisation of $(k+1)$ and the identity $k(2k+1)+6(k+1)=(k+2)(2k+3)$ into your book.

Did you get this? True or false: $(k+1)^2 = k^2 + 2k + 1$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FULL INDUCTION PROOF

Prove by induction that $1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all positive integers $n$.

Base case $n=1$: LHS $= 1$; RHS $= \dfrac{1 \cdot 2 \cdot 3}{6} = 1$. Equal. ✓

Substitute $n=1$ and evaluate both sides separately. Do not skip this step.

PROBLEM 2 · EVALUATING THE FORMULA

Evaluate $1^2 + 2^2 + 3^2 + \cdots + 10^2$.

Apply the formula with $n = 10$: $S_{10} = \dfrac{10 \times 11 \times 21}{6}$.

Identify $n = 10$. The three factors are $n = 10$, $n+1 = 11$, and $2n+1 = 21$.

PROBLEM 3 · FACTORISATION DRILL

In the inductive step for the sum of squares proof, the expression inside the bracket after factoring out $(k+1)$ is $k(2k+1) + 6(k+1)$. Show that this simplifies to $(k+2)(2k+3)$.

$k(2k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6$

Expand each term: $k(2k+1) = 2k^2+k$ and $6(k+1) = 6k+6$. Then collect like terms.

Fill the gap: $2k^2 + 7k + 6 = (k+2)($ $)$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Expanding $(k+1)^2$ incorrectly

$(k+1)^2 = k^2 + 2k + 1$. A very common error is writing $k^2 + 1$ (forgetting the cross term) or $k^2 + k + 1$ (only one cross term instead of two). If your quadratic doesn't come out to $2k^2+7k+6$, expand $(k+1)^2$ again carefully.

Trap 02

Missing the factor of 6

When combining $\dfrac{k(k+1)(2k+1)}{6} + (k+1)^2$, the second term must be written as $\dfrac{6(k+1)^2}{6}$ before combining numerators. Forgetting to multiply $(k+1)^2$ by $\dfrac{6}{6}$ is the single most common error in this proof.

Trap 03

Incorrect or unverified factorisation

If you write $2k^2+7k+6 = (k+3)(2k+2)$ by guessing, you will lose the mark. Always verify by expanding: $(k+2)(2k+3) = 2k^2+3k+4k+6 = 2k^2+7k+6$. Check against the target — the final line must be exactly $\dfrac{(k+1)(k+2)(2k+3)}{6}$.

Did you get this? True or false: when combining $\dfrac{k(k+1)(2k+1)}{6} + (k+1)^2$, the second term becomes $\dfrac{6(k+1)^2}{6}$ over the common denominator.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write out the complete induction proof for the sum of squares formula, showing all algebra in the inductive step.

Evaluate $1^2 + 2^2 + 3^2 + \cdots + 20^2$ using the formula.

Factorise $2k^2 + 7k + 6$ using the AC method, showing all working. Then verify by expanding your answer.

Without using the formula, find the value of $6^2 + 7^2 + 8^2 + 9^2 + 10^2$ by using the formula for $S_{10}$ and $S_5$.

Explain in your own words why factoring out $(k+1)$ before expanding makes the algebra easier in this proof.

Odd one out: Three of these are correct evaluations of the sum of squares formula. Which one is NOT?

Revisit your thinking

Earlier you directly added $1^2 + 2^2 + 3^2 + 4^2 + 5^2$ and estimated the 10-term sum.

The direct sum is $1+4+9+16+25 = \mathbf{55}$, confirmed by $\dfrac{5 \cdot 6 \cdot 11}{6} = 55$. The 10-term sum is $\dfrac{10 \cdot 11 \cdot 21}{6} = \mathbf{385}$. The key new skill in this proof is the quadratic factorisation — $2k^2+7k+6 = (k+2)(2k+3)$ — which you will encounter again in sum-of-cubes and higher power proofs.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 31 mark

Q1. Evaluate $1^2 + 2^2 + 3^2 + \cdots + 10^2$. (1 mark)

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ApplyBand 53 marks

Q2. Prove by mathematical induction that $1^2 + 2^2 + 3^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ for all positive integers $n$. (3 marks)

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AnalyseBand 52 marks

Q3. In the inductive step for the sum of squares proof, a student writes: $k(2k+1) + 6(k+1) = 2k^2 + 13k + 6$. Identify and correct the error. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers: 2. $S_{20} = \frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870$ · 4. $S_{10} - S_5 = 385 - 55 = 330$ · 3. AC: $ac=12$, $b=7$, split $3k+4k$; $2k^2+3k+4k+6 = k(2k+3)+2(2k+3)=(k+2)(2k+3)$; verify $(k+2)(2k+3)=2k^2+7k+6$ ✓.

Q1 (1 mark): $S_{10} = \dfrac{10 \times 11 \times 21}{6} = \dfrac{2310}{6} = \mathbf{385}$ [1].

Q2 (3 marks): Base case $n=1$: LHS $= 1$; RHS $= \frac{1 \cdot 2 \cdot 3}{6} = 1$. Equal [1]. Hypothesis: assume $S_k = \frac{k(k+1)(2k+1)}{6}$. Inductive step: $S_{k+1} = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{(k+1)[k(2k+1)+6(k+1)]}{6} = \frac{(k+1)(2k^2+7k+6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$ [1 for factorisation]. Conclusion by induction [1].

Q3 (2 marks): Error: the student used $k(2k+1) = 2k^2 + k$ but added $6 \times 2k = 12k$ instead of $6k$ from $6(k+1) = 6k+6$ [1]. Correct: $k(2k+1)+6(k+1) = 2k^2+k+6k+6 = 2k^2+7k+6$ [1].

Boss battle · The Proof Architect

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering sum of squares questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 5 · Induction for Geometric Series Lesson 7 · Induction for Sum of Cubes →

Module overview · Extension 1 · Checkpoint 2