已知数列An的通项公式An=2的n次方+n，求An的前n项和Sn

Question

如题所述

Answer 1

An=2^n+n
那么Sn=A1+A2+...+An
=(2^1+1)+(2^2+2)+...+(2^n+n)
=(2^1+2^2+...+2^n)+(1+2+...+n)
=2*(1-2^n)/(1-2)+n(n+1)/2
=2^(n+1)-2+n(n+1)/2

如果不懂，请追问，祝学习愉快！

Answer 2

an=2^n+n
sn=2^1+1+2^2+2+2^3+3+.......+2^n+n
=2^1+2^2+2^3+.......+2^n+1+2+3+.......+n
=2*(1-2^n)/(1-2)+n(n+1)/2
=2^(n+1)-2+n(n+1)/2
=2^(n+1)+(n^2+n-4)/2

Answer 3

额!这很难!追问

你会吗？很急啊！

追答

an=2^n+n
sn=2^1+1+2^2+2+2^3+3+.......+2^n+n
=2^1+2^2+2^3+.......+2^n+1+2+3+.......+n
=2*(1-2^n)/(1-2)+n(n+1)/2
=2^(n+1)-2+n(n+1)/2
=2^(n+1)+(n^2+n-4)/2