Does it ring a bell?

For $n\geq 1$ ; we have $(1\times2) +(2\times3)+(3\times4)+...+n(n+1) =$ $\frac{(n)(n+1)(n+2)}{3}$ ;

We can prove this by mathematical induction:

First the statement holds true for $n=1$ . The left hand side=2=Right hand side.

Suppose it holds true for all $n=k\in\ N$ ,

To show it is true for $k+1$ :

By taking the L.H.S, it will gives us:

$(1\times2)+(2\times3)+(3\times4)+...+k(k+1)+ (k+1)(k+1+1) =$

$\large(\frac{(k)(k+1)(k+2)}{3}$ )+ $(k+1)(k+2)$ =

$\large\frac{(k)(k+1)(k+2)}{3}$ + $\large\frac{3(k+1)(k+2)}{3}$ =

$\large\frac{(k+1)(k+2)(k+3)}{3}$ =

$\large\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$ $\implies$ our statement holds true for $(k+1)$ .

Then just take k=100 to get the answer.