Question

设数列{an}的前n项和为Sn=2n^2,{bn}为等比数列,且a1=b1,b2(a2--a1)=b1,设cn=an/b

Answer 1

Sn=2n^2

an=Sn-S(n-1)=4n-2

a1=b1=2,a2=6

b2(6-2)=b1

b2/b1=1/4

{bn}为等比数列

所以bn=2*(1/4)^(n-1)

cn=an/bn=(2n-1)*2^(2n-1)

Tn=c1+c2+……+cn

=2+3*2^3+5*2^5+……+(2n-3)*2^(2n-3)+(2n-1)*2^(2n-1)………………（1）式

4Tn=2^3+3*2^5+5*2^7+……+(2n-3)*2^(2n-1)+(2n-1)*2^(2n+1)…………（2）式

(1)-(2)得

-3Tn=2+2^4+2^6+……2^(2n)-(2n-1)*2^(2n+1)

Tn=10/9-[2^(2n+2)]/9+(2n-1)*2^(2n+1)/3