How to find the formula of the summation of i(i + 1) over 1 to N using Induction

Question

I am trying to solve a problem that requires me to find the formula for the summation of i(i + 1) from 1 to n (link to picture below) using Induction. I have no clue where to even start and I am a little frustrated. Any help is much appreciated.

Problem

Answer 1

Let g(x)=(1/3)(x+1)x(x-1). Then g(i+1)-g(i)=(i+1)i and thus $$\sum_{i=1}^N (i+1)i= \sum_{i=1}^N( g(i+1)-g(i)) $$ $$= g(N+1)-g(1) $$ $$=(1/3)(N+2)(N+1)N-(1/3)(2)(1)0 $$ $$=(N+2)(N+1)N/3 $$

Answer 2

Another way to do it without using the hint

Consider $$S_N=\sum_{i=1}^N (i+1)i x^i=\sum_{i=1}^N [i(i-1)+2i] x^i=\sum_{i=1}^N i(i-1) x^i+2\sum_{i=1}^N i x^i$$ Rewrite it as $$S_n=x^2\sum_{i=1}^N i(i-1) x^{i-2}+2x\sum_{i=1}^N i x^{i-1}=x^2 \left(\sum_{i=1}^N x^{i} \right)''+2x \left(\sum_{i=1}^N x^{i} \right)'$$ when finished, make $x=1$ (you will need to compute a limit).

Answer 3

The general case is $$\sum_{r=1}^nr(r+1)(r+2)\dots(r+k)=\frac{1}{k+2}n(n+1)(n+2)\dots(n+k+1)$$

For $n=1$ this is just: $$1\cdot2\cdot3\dots(k+1)=\frac{1}{k+2}1\cdot2\dots(k+2)$$ which is obviously true.

Suppose the result holds for $n$.

To establish it for $n+1$ we need to show that $$\frac{1}{k+2}(n+1)(n+2)\dots(n+k+2)=$$ $$\frac{1}{k+2}n(n+1)(n+2)\dots(n+k+1)+(n+1)(n+2)\dots(n+k+1)$$ Taking out the common factor $(n+1)\dots(n+k+1)$ from both sides, that is just

$$\frac{n+k+2}{k+2}=\frac{n}{k+2}+1$$ which is obviously true. So the result holds for $n+1$ and hence for all $n\ge1$.