Ta có :\(n^2-n=n.\left(n-1\right)\)
\(\implies\) \(n^2=\left(n-1\right)n+n\)
Áp dụng : với n=2017 thay vào ta có:
\(S=1+1.2+2+2.3+3+...+99.100+100\)
\(S=\left(1.2+2.3+...+99.100\right)+\left(1+2+...+100\right)\)
\(S=\frac{99.100.101}{3}+\frac{101.100}{2}\)
\(S=100.101.\left(\frac{99}{3}+\frac{1}{2}\right)\)
\(S=\frac{100.101.201}{6}\)