During the resolution of a physics problem, I have encountered the sum: $$\sum_{i=1}^{10} i(i+1)$$
For practical purposes, I evaluated it using a Python script, but a calculator would have been enough since the sum stops at $i=10$. However, I wonder what I could do mathematically with that sum if I did not have access to a computer and the sum reached large values of $i$.
Is there a way to evaluate it analytically?
There is the formula $$ \sum_{i=1}^ni(i+1)=\frac{1}{3}n(n+1)(n+2), $$ which you can prove by mathematical induction.
Observe that $$(i+1)^3 - i^3 - 1 = i^3 + 3i^2 + 3i + 1 - i^3 - 1 = 3(i^2 + i) = 3i(i+1).$$ Therefore, $$\begin{align} \sum_{i=1}^n i(i+1) &= \frac{1}{3}\sum_{i=1}^n \left((i+1)^3 - i^3 - 1\right) \\ &= \frac{1}{3} \left(\sum_{i=1}^n (i+1)^3 - \sum_{i=1}^n i^3 - \sum_{i=1}^n 1 \right) \\ &= \frac{1}{3} \left(\sum_{i=2}^{n+1} i^3 - \sum_{i=1}^n i^3\right) - \frac{n}{3} \\ &= \frac{1}{3} \left((n+1)^3 - 1^3 - n\right) \\ &= \frac{n^3 + 3n^2 + 2n}{3} \\ &= \frac{n(n+1)(n+2)}{3}. \end{align}$$
$$ \sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x} $$ $$ \sum_{i=0}^n x^{i+1} = x\frac{1-x^{n+1}}{1-x} $$ $$ \sum_{i=0}^n i(i+1)x^{i-1} = \frac{d^2}{dx^2}x\frac{1-x^{n+1}}{1-x} $$ $$ \sum_{i=0}^n i(i+1) = \frac{d^2}{dx^2}x\frac{1-x^{n+1}}{1-x} \; at\; x=1. $$
We have $$(i+1)^3-i^3=3i^2+3i+1\\=3i(i+1)+1$$ Thus$$\sum_{i=1}^ni(i+1)={1\over 3}\sum_{i=1}^n[(i+1)^3-i^3]+{1\over 3}n\\ ={1\over 3}[(n+1)^3-1]+{1\over 3}n$$
