Question

设数列｛an}的前n项和为sn，若对任意的正整数n都有sn=2an-3n,（1）设bn=an+3,求证

Answer 1

a1=S1=2a1-3,得a1=3
Sn=2an-3n
S(n-1)=2a(n-1)-3(n-1)
2式相减,得:
Sn-S(n-1)=2an-2a(n-1)-3=an
所以an+3=2(a(n-1)+3)
数列{an+3}是以a1+3=3+3=6为首项,以2为公比的等比数列
an+3=6*2^(n-1),an=3*2^n-3
bn=an+3=3*2^n
{bn}是以b1=6为首项,2为公比的等比数列.
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设数列{nan}的前n项和为Tn
则
Tn=3(1*2+2*2^2+3*2^3+...+n*2^n)-3n(n+1)/2
2Tn=3(
1*2^2+2*2^3+...+(n-1)*2^n+n*2^(n+1)-3n(n+1)
上式减去下式得:
-Tn=3(2+2^2+2^3+...+2^n)-3n*2^(n+1)+3n(n+1)/2
-Tn=3(2*(2^n-1))-3n*2^(n+1)+3n(n+1)/2
所以Tn=3(n-1)2^(n+1)-3n(n+1)/2+6