# proof by mathematical induction with the summation operator? [duplicate]

$$\sum_{k=1}^n k^3 = \left( \sum_{k=1}^n k \right)^2$$ I can't quite understand this expression, and in fact this is my biggest difficulty in finding a solution. Can someone please explain to me ? $$\sum_{k=1}^n k = \frac{n(n+1)} 2$$

• Left side is $1^3+2^3+\cdots+n^3$. Right side is $(1+2+\cdots+n)^2$ – paw88789 Nov 6 '14 at 18:47
• I love this result: it's quite simple and looks like it should be wrong. – Simon S Nov 6 '14 at 18:52
• – apnorton Nov 6 '14 at 19:13

It's asking you to prove that the following identity holds for all $n \in \mathbb N$: $$1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$$ For example, when $n = 5$, we have that: $$1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 = (1 + 2 + 3 + 4 + 5)^2$$

• can you prove by math induction please? – Matematika Matematika Nov 6 '14 at 18:49
• @MatematikaMatematika: have you used induction previously? What would be the base step? How would the assumption step look? – abiessu Nov 6 '14 at 18:50
• i guess i prove for n=1 then we assume for n=k it's true and then by using this assumption we prove it for n=k+1 – Matematika Matematika Nov 6 '14 at 18:51
• @MatematikaMatematika Have you already seen a closed form expression for $1 + 2 + \cdots + n$? That can help. – Adriano Nov 6 '14 at 18:52
• what about now?# – Matematika Matematika Nov 6 '14 at 18:56

You have $S_n=1+...+n = {1 \over 2} n (n+1)$. This is straightforward to show.

Let $C_n = 1^3+... +n^3$.

You want to show $C_n = S_n^2$. It is true for $n=1$, so suppose it is true for $n$.

Then $C_{n+1} = C_n + n^3$.

Also, $S_{n+1}^2 = (S_n + (n+1))^2 = S_n^2 +2 S_n (n+1) + (n+1)^2$.

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

Hence $C_{n+1} = S_{n+1}^2$.

you must prove that $1^3+2^3+...+n^3+(n+1)^3=(1+2+...+n+(n+1))^2$ and use that $1^3+2^3+...+n^3=(1+2+...+n)^2$ holds. Now you can write $1^3+2^3+3^3+...+n^3+(n+1)^3=(1+2+...+n)^2+(n+1)^3$ in the next step we have to show that $(1+2+...+n)^2+(n+1)^3=(1+2+...+n+(n+1))^2$

• what about now if we have the below expression – Matematika Matematika Nov 6 '14 at 18:56