Your weak spots

Insights load after your first practice round.

Module 5 · L8 of 20 ~40 min ⚡ +95 XP available

Induction for Higher Power Sums

You've proved $\sum r$, $\sum r^2$ and $\sum r^3$. Each formula is a polynomial one degree higher than the power summed. In this lesson you extend your induction skills to $\sum r^4$ — the algebra is harder, but the strategy is identical. You'll also discover a unifying pattern across all power sums.

Today's hook — We know $\sum r = \dfrac{n(n+1)}{2}$ (degree 2) and $\sum r^2 = \dfrac{n(n+1)(2n+1)}{6}$ (degree 3). Before we continue, predict: what degree polynomial should $\sum r^4$ be? And do you think the factor $n(n+1)$ still appears? Write your prediction below.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Write down the formulas you already know for $\sum r$, $\sum r^2$, and $\sum r^3$. From memory if possible. Then predict what degree polynomial $\sum r^4$ will be, and whether $n(n+1)$ will appear as a factor.

auto-saved

The pattern of power sums

+5 XP to read

We have proven three power sum formulas. Notice the degree pattern: $\sum r^m$ is always a polynomial of degree $m+1$.

Sum	Formula	Degree
$\displaystyle\sum_{r=1}^n r$	$\dfrac{n(n+1)}{2}$	2
$\displaystyle\sum_{r=1}^n r^2$	$\dfrac{n(n+1)(2n+1)}{6}$	3
$\displaystyle\sum_{r=1}^n r^3$	$\dfrac{n^2(n+1)^2}{4}$	4
$\displaystyle\sum_{r=1}^n r^4$	$\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$	5

$\deg\!\left(\sum r^m\right) = m+1$

Prove, don't derive

In the HSC you are given the formula and asked to prove it by induction. You do not need to find $\sum r^4$ from scratch.

$n(n+1)$ always factors

Every power sum formula contains $n(n+1)$ — one of $n$ or $n+1$ is always even, which is why the result is always an integer.

Verify first

Always check the formula for $n=1$ and $n=2$ before attempting the inductive step — it confirms you have the right formula.

What you'll master

Know

Key facts

$\displaystyle\sum_{r=1}^{n} r^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$
$\sum r^m$ is a polynomial of degree $m+1$ in $n$
All power sum formulas contain $n(n+1)$ as a factor

Understand

Concepts

Why the degree of $\sum r^m$ is one higher than $m$
How factoring out $\frac{(k+1)}{30}$ simplifies the inductive step for $\sum r^4$
Why the HSC tests proof by induction rather than derivation of formulas

Can do

Skills

Write a complete induction proof for $\sum r^4$, including careful factorisation
Evaluate $\sum r^4$ for specific values of $n$ using the formula
Identify the degree and factor structure of any power sum formula

Key terms

Sum of fourth powers$\displaystyle\sum_{r=1}^{n} r^4 = 1^4+2^4+\cdots+n^4$, a degree-5 polynomial in $n$.

Degree of a power sum$\sum r^m$ is a polynomial of degree $m+1$ in $n$. The pattern: $\sum r \to 2$, $\sum r^2 \to 3$, $\sum r^3 \to 4$, $\sum r^4 \to 5$.

Factor $n(n+1)$Every power sum formula contains $n(n+1)$; for odd powers (1, 3) only $n(n+1)$ appears; for even powers (2, 4) the extra factor $(2n+1)$ also appears.

Telescoping sumA technique to derive power sum formulas by summing $(r+1)^{m+1} - r^{m+1}$ and cancelling terms. Not required for the HSC but useful background.

Inductive step (fourth powers)Add $(k+1)^4$ to the sum for $n=k$, factor out $\frac{(k+1)}{30}$, expand and factorise the resulting cubic to reach the formula for $n=k+1$.

Algebraic persistenceHigher power proofs require expanded arithmetic. Always check the result against the formula for $n=2$ before submitting.

The fourth-power sum formula

core concept

For all positive integers $n$:

$$\sum_{r=1}^{n} r^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$

Verify for $n=1$: RHS $= \dfrac{1 \cdot 2 \cdot 3 \cdot 5}{30} = \dfrac{30}{30} = 1$ and LHS $= 1^4 = 1$. ✓

Verify for $n=2$: RHS $= \dfrac{2 \cdot 3 \cdot 5 \cdot 11}{30} = \dfrac{330}{30} = 11$ and LHS $= 1 + 16 = 17$. $\times$ — so check your arithmetic carefully before an exam. Actually: $3(4)+3(2)-1 = 12+6-1=17$, so $3n^2+3n-1\big|_{n=2} = 17$. RHS $= \dfrac{2 \cdot 3 \cdot 5 \cdot 17}{30} = \dfrac{510}{30} = 17$. ✓

Proof strategy: Factor out $\dfrac{(k+1)}{30}$ after adding $(k+1)^4$, then expand the remaining cubic and factorise it to match $(k+2)(2k+3)(3(k+1)^2+3(k+1)-1)$.

Why 30 in the denominator? The degree-5 polynomial $\sum r^4$ is always divisible by 30 because it contains the factors $n(n+1)(2n+1)$, which are three consecutive-ish integers guaranteeing divisibility by 2, 3 and 5. This is a useful sense-check.

Verify for $n=1$: RHS $= \dfrac{1 \cdot 2 \cdot 3 \cdot 5}{30} = \dfrac{30}{30} = 1$ and LHS $= 1^4 = 1$. ✓

Pause — copy the fourth-power sum formula $\sum_{r=1}^n r^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ and verify it for $n=1$ into your book.

Quick check: The formula $\displaystyle\sum_{r=1}^{n} r^4$ is a polynomial in $n$ of what degree?

The full induction proof

core concept

We just saw that $\sum_{r=1}^n r^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$, verified for $n=1$: $\frac{1\cdot2\cdot3\cdot5}{30}=1$ ✓. That raises a question: how do you add $(k+1)^4$ to the five-factor hypothesis and reassemble the numerator into $(k+2)(2k+3)(3k^2+9k+5)$? This card answers it → by expanding over denominator 30, factorising $(k+1)$, then matching coefficients in the cubic factor.

Claim: For all $n \geq 1$, $\displaystyle\sum_{r=1}^{n} r^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.

Step 1 — Base case ($n=1$):

LHS $= 1^4 = 1$. RHS $= \dfrac{1 \cdot 2 \cdot 3 \cdot (3+3-1)}{30} = \dfrac{1 \cdot 2 \cdot 3 \cdot 5}{30} = \dfrac{30}{30} = 1$. LHS $=$ RHS. ✓

Step 2 — Inductive hypothesis: Assume true for $n = k$:

$$\sum_{r=1}^{k} r^4 = \frac{k(k+1)(2k+1)(3k^2+3k-1)}{30}$$

Step 3 — Inductive step ($n=k+1$): We need to show:

$$\sum_{r=1}^{k+1} r^4 = \frac{(k+1)(k+2)(2k+3)(3(k+1)^2+3(k+1)-1)}{30}$$

Starting from LHS:

$= \dfrac{k(k+1)(2k+1)(3k^2+3k-1)}{30} + (k+1)^4$

$= \dfrac{(k+1)}{30}\left[k(2k+1)(3k^2+3k-1) + 30(k+1)^3\right]$

Expand the bracket step by step:

$k(2k+1)(3k^2+3k-1) = k(6k^3+9k^2+k-1) = 6k^4+9k^3+k^2-k$

$30(k+1)^3 = 30(k^3+3k^2+3k+1) = 30k^3+90k^2+90k+30$

Sum: $6k^4+39k^3+91k^2+89k+30$

Factorise: $= (k+2)(2k+3)(3k^2+9k+5)$

Note $3k^2+9k+5 = 3(k+1)^2+3(k+1)-1$, so:

$\displaystyle\sum_{r=1}^{k+1} r^4 = \dfrac{(k+1)(k+2)(2k+3)(3(k+1)^2+3(k+1)-1)}{30}$

This is the formula with $n = k+1$. Hence true for $n=k+1$. By induction, true for all $n \geq 1$. $\blacksquare$

Factorisation shortcut. After expanding $6k^4+39k^3+91k^2+89k+30$, look for the factor $(k+2)$ first (since the target formula requires it). Polynomial division gives $2k^3+35k^2+21k+15$... wait — try $(k+2)(2k+3)$ as a pair: $(k+2)(2k+3) = 2k^2+7k+6$. Long-dividing the quartic by this quadratic gives $3k^2+9k+5$.

Claim: For all $n \geq 1$, $\displaystyle\sum_{r=1}^{n} r^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.

Pause — copy the structure of the inductive step: expand numerator over 30, factor $(k+1)$, identify the target cubic $3k^2+9k+5$ into your book.

Did you get this? True or false: the first step of the inductive step is to factor $\dfrac{(k+1)}{30}$ from both terms.

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

PROBLEM 1 · VERIFY THE FORMULA

Find the value of $1^4+2^4+3^4+4^4+5^4$ using the formula.

Apply $\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ with $n=5$.

Identify the formula and substitute $n=5$.

PROBLEM 2 · DEGREE PATTERN

Without evaluating, state the degree of the polynomial $\displaystyle\sum_{r=1}^{n} r^6$ in $n$, and predict whether $n(n+1)$ will appear as a factor.

$\sum r^6$ has $m = 6$, so degree $= m+1 = 7$.

Apply the pattern: $\sum r^m$ is degree $m+1$.

PROBLEM 3 · INDUCTIVE STEP CHECK

In the induction proof for $\sum r^4$, verify that the polynomial $6k^4+39k^3+91k^2+89k+30$ equals $(k+2)(2k+3)(3k^2+9k+5)$ for $k=2$.

LHS at $k=2$: $6(16)+39(8)+91(4)+89(2)+30 = 96+312+364+178+30 = 980$

Substitute $k=2$ into the expanded polynomial.

Fill the gap: In the inductive step for $\sum r^4$, the expression $3k^2+9k+5$ can be rewritten as $3(k+1)^2 + 3(k+1) - $ , matching the required form for $n=k+1$.

Misconceptions to fix · 3 traps that cost marks

Trap 01

Expanding before factoring out $\frac{(k+1)}{30}$

If you expand $(k+1)^4$ into a degree-4 polynomial before extracting the common factor, you end up with a degree-5 expression with no clear path forward. Always factor $\dfrac{(k+1)}{30}$ immediately — this reduces the bracket to a cubic that can be factorised.

Trap 02

Assuming the algebra will factorise "magically"

Students sometimes write $(k+2)(2k+3)(3k^2+9k+5)$ without showing the expansion. In an exam worth 4 marks, you must show all working: expand the bracket, combine like terms, then factor. Skipping steps loses method marks.

Trap 03

Evaluating $3n^2+3n-1$ incorrectly for the base case

At $n=1$: $3(1)^2+3(1)-1 = 3+3-1 = 5$, not 6. A hasty student writes $3+3 = 6$ and gets RHS $= \dfrac{1 \cdot 2 \cdot 3 \cdot 6}{30} = \dfrac{36}{30} \neq 1$, causing them to abandon the proof. Always evaluate $3n^2+3n-1$ carefully before moving on.

Did you get this? True or false: the value of $3n^2+3n-1$ when $n=1$ is equal to 5.

Work mode · how are you completing this lesson?

Activities · practise the ideas

Verify the formula $\displaystyle\sum_{r=1}^{n} r^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ for $n=3$ by computing both sides.

Find $\displaystyle\sum_{r=1}^{4} r^4$ using the formula, and verify by direct computation.

State, without proof, the degree of $\displaystyle\sum_{r=1}^{n} r^8$ and predict whether $n(n+1)(2n+1)$ is a factor.

In the inductive step for $\sum r^4$, expand $k(2k+1)(3k^2+3k-1)$ and $30(k+1)^3$ separately, then add them.

Verify that $6k^4+39k^3+91k^2+89k+30 = (k+2)(2k+3)(3k^2+9k+5)$ by expanding the RHS.

Odd one out: Three of these statements about $\displaystyle\sum_{r=1}^{n} r^m$ are true. Which one is FALSE?

Revisit your thinking

Earlier you predicted the degree of $\sum r^4$ and whether $n(n+1)$ appears.

The degree is 5 — one higher than the power 4, consistent with the pattern for all power sums. And yes, $n(n+1)$ always appears: the formula $\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ contains it explicitly. The hard part of this lesson was the inductive step's lengthy factorisation — but the strategy is always the same: extract the common factor of the next term first, then deal with what remains.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

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. Find the exact value of $1^4+2^4+3^4+4^4+5^4$. (1 mark)

auto-saved

ApplyBand 54 marks

Q2. Prove by mathematical induction that $\displaystyle\sum_{r=1}^{n} r^4 = \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ for all positive integers $n$. (4 marks)

auto-saved

AnalyseBand 52 marks

Q3. State the degree of $\displaystyle\sum_{r=1}^{n} r^5$ and explain, using the pattern from power sums, why you would expect $n(n+1)$ to be a factor of the formula. (2 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $n=3$: $3n^2+3n-1 = 27+9-1 = 35$. RHS $= \dfrac{3 \cdot 4 \cdot 7 \cdot 35}{30} = \dfrac{2940}{30} = 98$. LHS $= 1+16+81 = 98$ ✓

2. $n=4$: $3(16)+12-1=59$. RHS $= \dfrac{4 \cdot 5 \cdot 9 \cdot 59}{30} = \dfrac{10620}{30} = 354$. Direct: $1+16+81+256 = 354$ ✓

3. Degree $= 9$; yes, $n(n+1)$ appears; for even power 8, $(2n+1)$ is also expected.

4. $k(2k+1)(3k^2+3k-1) = 6k^4+9k^3+k^2-k$; $30(k+1)^3 = 30k^3+90k^2+90k+30$; sum $= 6k^4+39k^3+91k^2+89k+30$.

5. $(k+2)(2k+3) = 2k^2+7k+6$; $(2k^2+7k+6)(3k^2+9k+5) = 6k^4+39k^3+91k^2+89k+30$ ✓

Q1 (1 mark): $\dfrac{5 \cdot 6 \cdot 11 \cdot 89}{30} = \dfrac{29370}{30} = \mathbf{979}$ [1].

Q2 (4 marks): Base $n=1$: LHS $=1$, RHS $=30/30=1$ [1]. Hypothesis: assume $\sum_{r=1}^k r^4 = \dfrac{k(k+1)(2k+1)(3k^2+3k-1)}{30}$ [0]. Step: $\dfrac{(k+1)}{30}\left[k(2k+1)(3k^2+3k-1)+30(k+1)^3\right]$ [1]; expand bracket to $6k^4+39k^3+91k^2+89k+30$ [1]; factor as $(k+2)(2k+3)(3k^2+9k+5)$ and note $3k^2+9k+5=3(k+1)^2+3(k+1)-1$ to reach required form [1]. Conclude by induction [0].

Q3 (2 marks): Degree $= 5+1 = 6$ [1]. Every power sum $\sum r^m$ contains $n(n+1)$ because consecutive integers always include one multiple of 2; this factor pair is universal. For odd $m$, only $n(n+1)$ is needed; for even $m$, $(2n+1)$ also appears [1].

Boss battle · The Power Master

earn bronze · silver · gold

Five timed questions on higher power sums and induction. 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 higher power sum questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 7 · Sum of Cubes Lesson 9 · Divisibility I →

Module overview · Extension 1 · Checkpoint 2