Em chi lam bua thui :3
\(S_n=C^3_3+C^3_4+...+C^3_n\)
\(C^3_n=\dfrac{n!}{3!\left(n-3\right)!}=\dfrac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{6.\left(n-3\right)!}=\dfrac{n\left(n-1\right)\left(n-2\right)}{6}\)
\(\Rightarrow S_n=\dfrac{1.2.3}{6}+\dfrac{2.3.4}{6}+....+\dfrac{n\left(n-1\right)\left(n-2\right)}{6}\)
\(S_n=\dfrac{1}{6}\left(1.2.3+2.3.4+...+n\left(n-1\right)\left(n-2\right)\right)\)
Mục đích là tính tổng \(S_n'=1.2.3+....+n\left(n-1\right)\left(n-2\right)\)
\(4S'_n=1.2.3.4+2.3.4.4+...+4n\left(n-1\right)\left(n-2\right)\)
\(4S_n'=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+n\left(n-1\right)\left(n-2\right)\left[\left(n+1\right)-\left(n-3\right)\right]\)
\(\Rightarrow S'_n=\dfrac{n\left(n-1\right)\left(n-2\right)\left(n+1\right)}{4}\)
Den day tim lim la duoc =))
Dung ko nhi :D?
