Mathematical Induction Problem solving

Question

$1^3$ + $2^3$ + $2^3$ + ... + $n^3$ = ($1 + 2 + 3 + ... + n)^2$

I start with $P(1)$ and get $1 = 1$.

Then I do it with $P(n+1)$ and I get stuck.

$1^3$ + $2^3$ + $2^3$ + ... + $n^3$ + $(n+1)^3$ = ($1 + 2 + 3 + ... + n +(n+1))^2$

then I've tried substituting values and both ways and I cannot find anywhere to go with the problem.

$(1 + 2 + 3 + ... + n)^2 + (n+1)^3 = (1 + 2 + 3 + ... + n +(n+1))^2$ $OR$

$1^3 + 2^3 + 2^3 + ... + n^3 + (n+1)^3 = (1^3 + 2^3 + 2^3 + ... + n^3 +(n+1))^2$

I know that the sum of numbers in a row is $\cfrac{n(n+1)}{2}$. I'm not sure if that's of any use. I'm pretty sure there's a mathematical proof that I can't remember that'll clean up this problem so any insight would be greatly appreciated.

Answer 1

The fact that $\sum_{i=1}^n i=n(n+1)/2$ is of use! You need to use it no matter which side you choose first. If we take the LHS first we have:

$$ \begin{split} 1^3+2^3+\dots +n^3+(n+1)^3 &= (1+2+\dots +n)^2+(n+1)^3 \\ &=\left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\&= \frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} \\ &= \frac{(n+1)^2[n^2+4(n+1)]}{4} \\ &=\frac{(n+1)^2(n+2)^2}{4} \\ &=\left( \frac{(n+1)(n+2)}{2}\right)^2 \\ &= (1+2+\dots +n+(n+1))^2 = RHS \end{split} $$

as required. I think starting with the RHS is slightly simpler.

Answer 2

Suppose $\left(\sum_{k=1}^n k\right)^2 =\sum_{k=1}^n k^3 $.

Then

$\begin{array}\\ \left(\sum_{k=1}^{n+1} k\right)^2 &=\left(\sum_{k=1}^{n} k+(n+1)\right)^2\\ &=\left(\sum_{k=1}^n k\right)^2+2(n+1)\left(\sum_{k=1}^{n} k\right)+(n+1)^2\\ &=\sum_{k=1}^n k^3+2(n+1)\left(\sum_{k=1}^{n} k\right)+(n+1)^2 \qquad\text{(by the induction hypothesis)}\\ \end{array} $

So, we need to show that $(n+1)^3 =2(n+1)\left(\sum_{k=1}^{n} k\right)+(n+1)^2 $, because this will show that the right side is $\sum_{k=1}^{n+1} k^3 $.

Dividing by $n+1$, this is $(n+1)^2 =2\left(\sum_{k=1}^{n} k\right)+(n+1) $, or $\sum_{k=1}^{n} k =\frac{(n+1)^2-(n+1)}{2} =\frac{n(n+1)}{2} $.

This you should know or can easily prove by induction.