Induction for Sum of Cubes
The sum of the first $n$ cubes hides a stunning secret: it equals the square of the sum of the first $n$ integers. You can check it by hand for small $n$ — but can you prove it holds forever? In this lesson you'll write the cleanest possible induction proof and unlock one of the most beautiful identities in all of mathematics.
Compute $1^3 + 2^3 + 3^3$ and compare it to $(1+2+3)^2$. Do not look up the formula yet — just crunch the numbers. Write your two values and a prediction about whether the pattern continues for $n = 4$.
The sum of the cubes of the first $n$ positive integers is:
The closed-form formula is:
$$1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$$
This can be rewritten as $\left[\dfrac{n(n+1)}{2}\right]^2$, which is exactly $\left[\displaystyle\sum_{r=1}^{n} r\right]^2$ — the square of the sum of the first $n$ integers. That is the beautiful secret: the sum of cubes is the square of the sum of integers.
Key facts
- $\displaystyle\sum_{r=1}^{n} r^3 = \dfrac{n^2(n+1)^2}{4}$ for all $n \geq 1$
- This equals $\left(\displaystyle\sum_{r=1}^{n} r\right)^2$
- The inductive step requires factoring out $(k+1)^2$
Concepts
- Why the sum of cubes equals the square of the triangular number
- How factoring $(k+1)^2$ simplifies the algebra dramatically
- Why recognising $k^2+4k+4=(k+2)^2$ is the key finishing move
Skills
- Write a complete, rigorous induction proof for the sum of cubes
- Verify the formula by direct calculation for small values of $n$
- Explain the connection between $\sum r^3$ and $(\sum r)^2$
For all positive integers $n$:
An equivalent and elegant form uses the triangular number $T_n = \dfrac{n(n+1)}{2}$:
This is the Nicomachus theorem, known since antiquity. The proof by induction uses three key moves:
- Use the inductive hypothesis to replace the sum to $k$.
- Factor out $(k+1)^2$, which appears in both terms.
- Recognise $k^2 + 4k + 4 = (k+2)^2$ to reach the target form.
Sum of cubes: $1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}=(T_n)^2$ where $T_n=\frac{n(n+1)}{2}$ is the $n$th triangular number.
Pause — copy the sum-of-cubes formula $\sum_{r=1}^n r^3 = \frac{n^2(n+1)^2}{4}$ and the equivalent form $(T_n)^2$ into your book.
Quick check: The sum $1^3 + 2^3 + 3^3$ equals which of the following?
We just saw that $1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}=(T_n)^2$ where $T_n$ is the $n$th triangular number. That raises a question: how do you factor $\frac{k^2(k+1)^2}{4}+(k+1)^3$ to match $\frac{(k+1)^2(k+2)^2}{4}$? This card answers it → factorise $(k+1)^2$ and note $k^2+4(k+1)=(k+2)^2$.
Claim: For all $n \geq 1$, $\displaystyle\sum_{r=1}^{n} r^3 = \dfrac{n^2(n+1)^2}{4}$.
Step 1 — Base case ($n=1$):
LHS $= 1^3 = 1$. RHS $= \dfrac{1^2 \cdot 2^2}{4} = \dfrac{4}{4} = 1$. LHS $=$ RHS. ✓
Step 2 — Inductive hypothesis: Assume the statement is true for $n = k$, i.e.
Step 3 — Inductive step ($n = k+1$): We need to show:
Starting from LHS:
$\text{LHS} = \dfrac{k^2(k+1)^2}{4} + (k+1)^3$ (by the inductive hypothesis)
$= \dfrac{k^2(k+1)^2 + 4(k+1)^3}{4}$ (common denominator)
$= \dfrac{(k+1)^2\left[k^2 + 4(k+1)\right]}{4}$ (factor out $(k+1)^2$)
$= \dfrac{(k+1)^2(k^2 + 4k + 4)}{4}$ (expand bracket)
$= \dfrac{(k+1)^2(k+2)^2}{4}$ (recognise perfect square) $= \text{RHS}$
Hence if the statement holds for $n=k$, it holds for $n=k+1$. By the principle of mathematical induction, the statement is true for all $n \geq 1$. $\blacksquare$
Claim: For all $n \geq 1$, $\displaystyle\sum_{r=1}^{n} r^3 = \dfrac{n^2(n+1)^2}{4}$.
Pause — copy the inductive step, highlighting the factorisation $(k+1)^2[k^2+4(k+1)]=(k+1)^2(k+2)^2$ into your book.
Did you get this? True or false: the key algebraic move in the inductive step is factoring out $(k+1)^2$ from both terms of the numerator.
Worked examples · 3 in a row, reveal as you go
Verify that $1^3 + 2^3 + 3^3 + 4^3 = (1+2+3+4)^2$.
Find the exact value of $1^3 + 2^3 + \cdots + 6^3$.
In the proof by induction for $\sum r^3 = \dfrac{n^2(n+1)^2}{4}$, show explicitly that $\dfrac{k^2(k+1)^2}{4} + (k+1)^3 = \dfrac{(k+1)^2(k+2)^2}{4}$.
Fill the gap: In the inductive step, after factoring out $(k+1)^2$, the bracket simplifies to $k^2 + 4k + 4 = $ .
Misconceptions to fix · 3 traps that cost marks
Did you get this? True or false: $(1+2+3+4+5)^2 = 225$, and this equals $1^3+2^3+3^3+4^3+5^3$.
Activities · practise the ideas
Verify the formula $\dfrac{n^2(n+1)^2}{4}$ for $n = 5$ by computing both LHS and RHS directly.
Find $\displaystyle\sum_{r=1}^{10} r^3$ using the formula.
Write out the inductive step in full, starting from $\dfrac{k^2(k+1)^2}{4} + (k+1)^3$ and reaching $\dfrac{(k+1)^2(k+2)^2}{4}$.
Use the connection $\sum r^3 = T_n^2$ to evaluate $1^3+2^3+\cdots+8^3$ without expanding any cubes.
If $\displaystyle\sum_{r=1}^{n} r^3 = 3025$, find $n$. (Hint: $\sqrt{3025} = T_n$ for some $n$.)
Odd one out: Three of these are equivalent to $\displaystyle\sum_{r=1}^{n} r^3$. Which one is NOT?
Earlier you computed $1^3+2^3+3^3$ and compared it to $(1+2+3)^2$, and made a prediction about $n=4$.
You now know the result is $\textbf{not a coincidence}$ — it is a theorem. The key reason: the triangular number $T_n = \dfrac{n(n+1)}{2}$ satisfies $T_n^2 = \sum r^3$ for every $n$. The induction proof shows exactly why adding the next cube $(k+1)^3$ preserves this square structure.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find $\displaystyle\sum_{r=1}^{7} r^3$. (1 mark)
Q2. Prove by induction that $1^3 + 2^3 + \cdots + n^3 = \dfrac{n^2(n+1)^2}{4}$ for all positive integers $n$. (3 marks)
Q3. Explain, using the identity $\displaystyle\sum_{r=1}^{n} r^3 = T_n^2$, why $1^3+2^3+\cdots+n^3$ is always a perfect square for any positive integer $n$. (2 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\dfrac{5^2 \cdot 6^2}{4} = \dfrac{900}{4} = 225$; also $(1+2+3+4+5)^2 = 15^2 = 225$ ✓
2. $\dfrac{10^2 \cdot 11^2}{4} = \dfrac{12100}{4} = 3025$ · 3. (see card 06 for full working) · 4. $T_8 = 36$, so $\sum r^3 = 36^2 = 1296$ · 5. $\sqrt{3025} = 55 = T_n$; $\dfrac{n(n+1)}{2} = 55 \Rightarrow n(n+1) = 110 \Rightarrow n = 10$
Q1 (1 mark): $\dfrac{7^2 \cdot 8^2}{4} = \dfrac{49 \cdot 64}{4} = \dfrac{3136}{4} = \mathbf{784}$ [1]. Check: $T_7 = 28$, $28^2 = 784$ ✓
Q2 (3 marks): Base: $n=1$: LHS $=1$, RHS $=\tfrac{4}{4}=1$ [1]. Hypothesis: assume $\sum_{r=1}^k r^3 = \tfrac{k^2(k+1)^2}{4}$ [0.5]. Step: $\tfrac{k^2(k+1)^2}{4} + (k+1)^3 = \tfrac{(k+1)^2(k^2+4k+4)}{4} = \tfrac{(k+1)^2(k+2)^2}{4}$ [1.5]. Conclusion by induction [0]. (Marks: 1 base + 2 step.)
Q3 (2 marks): $T_n = \dfrac{n(n+1)}{2}$ is always a positive integer for any $n \geq 1$ (since $n$ or $n+1$ is even, the fraction is an integer) [1]. Since $\sum r^3 = T_n^2$, it is the square of an integer, hence always a perfect square [1].
Five timed questions on sum of cubes and the induction proof. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering sum-of-cubes questions. Lighter alternative to the boss.
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