Prove that $\left(\sum_{k=1}^{n}k\right)^2=\sum_{k=1}^{n}k^3$ holds true for $n ≥ 1$

Question

I've been trying to figure out this proof for way too long now, I'm just not sure where to begin for the inductive step. Any guidance would be greatly appreciated.

Answer 1

Hint:

$$ (1+2+\dots+n + (n+1))^2 - (1+2+\dots+n )^2 \\= (n+1)(2\times 1+2\times2+\dots+2\times n + (n+1)) $$

now remember that $$ 1+2+\dots+n = \frac12 n(n+1) $$

Answer 2

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

Answer 3

Suppose it holds for some $n \geq 1$.

Claim: It holds for $n+1$, too.

Proof: We have

\begin{align*} \left(1+ 2 + \ldots + n + (n+1) \right)^2 & = \left(1 + 2 + \ldots + n \right)^2 + 2 \left(1 + 2 + \ldots + n \right) (n+1) + (n+1)^2 \\ & = (1^3 + 2^3 + \ldots + n^3) + 2\frac{n(n+1)}{2} (n+1) + (n+1)^2 \\ & = (1^3 + 2^3 + \ldots + n^3) + n(n+1) ^2 + (n+1)^2 \\ & = 1^3 + 2^3 + \ldots + n^3 + (n+1)^3 \end{align*}

(I have made use of Gauss's addition forumla)

The assertion holds trivially for $n = 1$, it follows from induction that it holds for all $n \geq 1$.

Answer 4

To make this proof easier to follow, let's define the following: $$\begin{align*} A_n &= \sum_{k=1}^n k, \\ B_n &= A_n^2 = \biggl(\sum_{k=1}^n k\biggr)^2, \\ C_n &= \sum_{k=1}^n k^3. \end{align*}$$ Then the claim to be proven is that $B_n = C_n$ for all positive integers $n$. Here, we focus on the inductive step: that is, we wish to establish that if there exists a positive integer $n$ such that $B_n = C_n$, this implies that $B_{n+1} = C_{n+1}$. To this end, we first observe $A_{n+1} = (n+1) + A_n$: hence $$\begin{align*} B_{n+1} &= A_{n+1}^2 = ((n+1) + A_n)^2 \\ &= (n+1)((n+1) + 2A_n) + A_n^2 \\ &= (n+1)(A_n + (n+1) + A_n) + B_n \\ &= (n+1)(A_{n+1} + A_n) + C_n. \end{align*}$$ But note $$A_{n+1} + A_n = (n+1) + \sum_{k=1}^n k + \sum_{k=n}^1 ((n+1)-k) = (n+1) + \sum_{k=1}^n (n+1) = (n+1)^2,$$ thus $$B_{n+1} = (n+1)^3 + C_n = C_{n+1},$$ which completes the inductive step.

Answer 5

Hint: We know that $\sum_{k=1}^nk={n(n+1)\over 2}$.