Question

已知数列{An}的钱n项和为n(n+1),而数列｛Bn｝的第n项Bn等于数列{An}的第2^n项,即Bn=A2^n.

Answer 1

1.设An前n项和为Tn,T1=A1=2,n大于等于2时,An=Tn-T(n-1)=n^2+n-n^2-n=2n,n=1时也成立,所以An=2n,Bn=A2^n=2^(n+1)

2.由等比数列前n项和,Sn=4(2^n-1)=4*2^n-4