题目

已知数列｛an｝中,a1=2,a↓n+1=2an+3 1）求an .2）令bn=n an,求数列｛bn｝的前n项和sn

提问时间：2020-11-23

答案

a(n+1)=2an+3
a(n+1)+3=2an+6
a(n+1)+3=2(an+3)
[a(n+1)+3]/(an+3)=2
所以an+3是等比数列,公比为2
an+3=(a1+3)q^(n-1)
an+3=(2+3)*2^(n-1)
an=5*2^(n-1)-3
bn=n an
bn=n[5*2^(n-1)-3]
bn=5n*2^(n-1)-3n
sn=5*1*2^(1-1)-3*1+5*2*2^(2-1)-3*2+.+5n*2^(n-1)-3n
=5*[1*2^(1-1)+2*2^(2-1)+.+n*2^(n-1)]-3(1+2+3+...+n)
=5*[1*2^(1-1)+2*2^(2-1)+.+n*2^(n-1)]-3n(n+1)/2
2sn=5*[1*2^(2-1)+2*2^(3-1)+.+n*2^n]-3n(n+1)
sn-2sn=5*[2^(1-1)+2^(2-1)+.+2^(n-1)-n*2^n]-3n(n+1)/2+3n(n+1)
-sn=5*[(1-2^n)/(1-2)-n*2^n]+3n(n+1)/2
-sn=5*[2^n-1-n*2^n]+3n(n+1)/2
sn=5*[n*2^n-2^n+1]-3n(n+1)/2
sn=5*[(n-1)*2^n+1]-3n(n+1)/2
sn=5(n-1)*2^n-3n(n+1)/2+5