Question

\(A=C^0_{2000}+2C^1_{2000}+3C^2_{2000}+...+2001C^{2000}_{2000}\)

Rút gọn KHÔNG DÙNG ĐẠO HÀM, TÍCH PHÂN

Answer 1

\(\left(k+1\right)C^k_n=kC^k_n+C^k_n=\dfrac{n!k}{k!\left(n-k\right)!}+C^k_n=\dfrac{\left(n-1\right)!n}{\left(k-1\right)!\left(n-1-k+1\right)!}+C^k_n=nC^{k-1}_{n-1}+C^k_n\)

\(\Rightarrow C^0_{2000}+\sum\limits^{2000}_{k=1}\left(k+1\right)C^k_{2000}=C^0_{2000}+\sum\limits^{2000}_{k=1}\left(2000C^{k-1}_{1999}+C^k_{2000}\right)=2000\sum\limits^{2000}_{k=1}C^{k-1}_{1999}+\sum\limits^{2000}_{k=0}C^k_{2000}\)

\(=2000.2^{1999}+2^{2000}=2^{1999}.2002\)