Problems have identity crisis too!

I'll use the identity ${n \choose 2} + {n+1 \choose 2} = n^2$ Now, $\xi$ can be expressed as follows ${\large \begin{aligned} \xi & = \sum_{n=2}^{100} {n \choose 2} \\ & = {2 \choose 2}+{3 \choose 2}+{4 \choose 2}+{5 \choose 2}+{6 \choose 2}+{7 \choose 2}+ \ldots + {98 \choose 2}+{99 \choose 2}+{100 \choose 2} \\ & = 2^2+4^2+6^2+\ldots+98^2+{100 \choose 2} \\ & = 2^2 (1^2+2^2+3^2 + \ldots + 49^2) +{100 \choose 2} \\ & = 4 \frac{49 \times 50 \times 99}{6} + \frac{99 \times 100}{2} \\ & = 166650 \end{aligned} }$

As for the proof of the identity ${\large \begin{aligned} {n \choose 2} + {n+1 \choose 2} & = \frac{n!}{2!(n-2)!} +\frac{(n+1)!}{2!(n-1)!} \\ & = \frac{n!(n-1)}{2!(n-1)!} + \frac{(n+1)!}{2!(n-1)!} \\ & = \frac{n!(n-1) +(n+1)!}{2!(n-1)!} \\ & = \frac{n!(n-1+n+1)}{2!(n-1)!} \\ & = \frac{2n^2}{2!} \\ & = n^2 \end{aligned} }$

Applying the hockey stick identity $\displaystyle \sum_{n=k}^{N} \dbinom{n}{k} = \dbinom{N + 1}{k + 1}$ with $k = 2, N = 100$ , we find that $\xi = \dbinom{101}{3} = \boxed{166650}$ .