Question

已知数列{an}前N项和为Sn=(2^n+1)减2（1）求an=?(2)设bn=an^2,求数列{bn}的前N项和Tn=

Answer 1

1.

sn=2^(n+1)-2=2*2^n-2

->s(n-1)=2^n-2

->an=sn-s(n-1)=2^n

2.

bn=an^2=2^n^2=2^2n=4^n

Tn=4*(-1+4^n)/3