Exercise 1043Logic, sets and proofs › Proof techniques › InductionL1Added 3 Nov 2026

Sum of cubes

#induction#sum of cubes

Statement

Prove by induction: $\sum_{k=1}^n k^3=\left(\frac{n(n+1)}{2}\right)^2$ .

Hint

Note that $1^3+\cdots+n^3=(1+\cdots+n)^2$ .

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Solution

Base case. $n=1$ : $1=1^2$ . ✓

Inductive step. Assume the identity. Then
$\sum_{k=1}^{n+1}k^3=\frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{(n+1)^2[n^2+4(n+1)]}{4}=\frac{(n+1)^2(n+2)^2}{4}.$
That's $\left(\frac{(n+1)(n+2)}{2}\right)^2$ .

Conclusion. The formula holds.

Sum of cubes

Statement

Solution

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