已知数列{an}的通项公式为an=2n-2^n,求它的前n项和Sn

2025-06-26 21:48:40
推荐回答(1个)
回答1:

Sn=2*(1+2+3+...+n)-(2+2^2+2^3+...+2^n)
=2*(1+n)*n/2-(2*(1-2^n)/(1-2))
=n(n+1)-2(2^n-1)
=n(n+1)-2^(n+1)+2