Proof by induction verification

Question

I think I'm using what I'm trying to prove in the induction step. To prove is $\sum_{k=1}^nk(n-k)=\frac{(n-1)n(n+1)}{6}$. My induction step is $$\sum_{k=1}^{n+1}k\left((n+1)-k\right)=\underbrace{\sum_{k=1}^nk\left((n+1)-k\right)+\sum_{k=n+1}^{n+1}k(n-k)}_{\text{break the sum in 2 sums}}=\underbrace{\left[\frac{(n+1-1)(n+1)(n+1+1)}{6}\right]}_{\text{we supposed it holds for n}}+\underbrace{0}_{\text{since }k=n}.$$ Rearrange the terms in the part that doesn't vanish $$\frac{(n+1-1)(n+1)(n+1+1)}{6}=\frac{\left((n+1)-1\right)(n+1)\left((n+1)+1\right)}{6}$$ is this correct?

Answer 1

You know that $\sum_{k=1}^nk(n-k)=\frac{(n-1)n(n+1)}{6}$, not that $\sum_{k=1}^nk((n+1)-k)=\frac{((n+1)-1)(n+1)((n+1)+1)}{6}$, so your proof is wrong. You have to decompose the sum even more. So:

$$\sum_{k=1}^{n+1}k((n+1)-k) = \sum_{k=1}^{n}k((n+1)-k) = \sum_{k=1}^{n}k(n-k) + \sum_{k=1}^{n}k = \frac{(n-1)n(n+1)}{6} + \frac{n(n+1)}{2}$$ $$\frac{n(n+1)(n-1+3)}{6} = \frac{((n+1)-1)(n+1)((n+1) + 1)}{6}$$

Hence the proof.