Question

Giúp em bài 5 với 🙏🙏🙏

Answer 1

\(\dfrac{k.C_n^k}{C_n^{k-1}}=\dfrac{k.n!}{k!\left(n-k\right)!}.\dfrac{\left(k-1\right)!.\left(n-k+1\right)!}{n!}=n-k+1\)

Do đó:

\(S=2016-0+2016-1+...+2016-2015\)

\(=2016+2015+...+1=\dfrac{2016.2017}{2}=...\)