Prove by induction that $\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$

Question

Show the following for all positive integers using proof by induction:

$$\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$$

Base case (n = 1) passes: $1^3 = 1^2$

We assume the following: $$\sum_{k=1}^{p} k^3 = \bigg( \sum_{k=1}^{p}k\bigg)^2$$

This gives us:

$$\sum_{k=1}^{p+1} k^3 = \bigg( \sum_{k=1}^{p}k\bigg)^2 + (p+1)^3 = \bigg( \sum_{k=1}^{p+1}k\bigg)^2$$

This is as far as I have gotten. I have no Idea as to how I expand it so it becomes obvious that they are in fact equal.

Answer 1

We have $$\Big(\sum_{k=1}^{p+1} k\Big)^2=\Big(\sum_{k=1}^p k + (p+1)\Big)^2 = \Big(\sum_{k=1}^p k\Big)^2 + \color{blue}{2 (p+1)\Big(\sum_{k=1}^p k\Big)+(p+1)^2}.$$

Since $\sum_{k=1}^p k = \frac12 p(p+1)$, $$\color{blue}{2(p+1)\Big(\sum_{k=1}^p k\Big)+(p+1)^2} = p(p+1)^2+(p+1)^2=(p+1)^3.$$

Answer 2

A hint is to write $$\left(p + \sum_{k=1}^{p-1}k\right)^2$$and then apply binomial theorem when treating the sum as an already known number (it is an arithmetic sum).

Answer 3

We have $$\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$$ 1. Putting $n=1$ in the above equality, we get $$\sum_{k=1}^{1} k^3 = \bigg( \sum_{k=1}^{1}k\bigg)^2$$ $$ (1)^3=(1)^2\iff 1=1$$ Hence, the equality holds for $n=1$

2. Assuming that it holds for $n=p$ then we have $$\sum_{k=1}^{p} k^3 = \bigg( \sum_{k=1}^{p}k\bigg)^2$$ $$\color{blue}{\sum_{k=1}^{p} k^3=\left(\frac{p(p+1)}{2}\right)^2}$$

3. Putting $n=p+1$, we get
   $$\sum_{k=1}^{p+1} k^3 = \bigg( \sum_{k=1}^{p+1}k\bigg)^2$$ $$\sum_{k=1}^{p} k^3+(p+1)^3=\left(\frac{(p+1)(p+2)}{2}\right)^2$$ $$\sum_{k=1}^{p} k^3=\left(\frac{(p+1)(p+2)}{2}\right)^2-(p+1)^3$$

$$=(p+1)^2\left(\frac{(p+2)^2}{4}-(p+1)\right)$$

$$=(p+1)^2\left(\frac{p^2+4p+4-4p-4}{4}\right)$$ $$=(p+1)^2\left(\frac{p^2}{4}\right)$$ $$=\left(\frac{p^2(p+1)^2}{4}\right)$$ $$=\left(\frac{p(p+1)}{2}\right)^2 $$ Which is true from (2)

Hence the equality holds for all positive integers $\color{blue}{n\geq 1}$

Answer 4

It comes to the same as proving that:

1) $\sum_{k=1}^{1}k^{3}=\left(\sum_{k=1}^{1}k\right)^{2}$ (trivial) and:

2) $\left(\sum_{k=1}^{n}k\right)^{2}-\left(\sum_{k=1}^{n-1}k\right)^{2}=n^{3}$ for $n=2,3\dots$

Note that the LHS equals:

$\left[\sum_{k=1}^{n}k+\sum_{k=1}^{n-1}k\right]\left[\sum_{k=1}^{n}k+\sum_{k=1}^{n-1}k\right]=n\left(2\sum_{k=1}^{n}k-n\right)$

and that:

$2\sum_{k=1}^{n}k=\sum_{k=1}^{n}k+\sum_{k=1}^{n}\left(n-k+1\right)=\sum_{k=1}^{n}\left(n+1\right)=n\left(n+1\right)$