51问答网 > 已知数列{an}的通项an=(1⼀2)^n+3n-2,求前n项和Sn

已知数列{an}的通项an=(1⼀2)^n+3n-2,求前n项和Sn

2025-03-11 06:31:11

推荐回答（1个）

回答1：

an=(1/2)^n+3n-2

Sn=a1+a2+a3+............+an
=(1/2)^1+3*1-2+(1/2)^2+3*2-2+(1/2)^3+3*3-2+..........+(1/2)^n+3n-2
=(1/2)^1+(1/2)^2+(1/2)^3+..........+(1/2)^n+3n-2+3(1+2+3+......+n)-2n
=1/2*[1-(1/2)^n]/(1-1/2)+3n(n+1)/2-2n
=1-(1/2)^n+(3n^2+3n/2-2n
=-(1/2)^n+(3n^2+3n-4n+2)/2
=-(1/2)^n+(3n^2-n+2)/2