Question

已知数列an的通项公式为an=n^2,数列bn的首项为3,其前n项和为sn,且满足（7n+1）/6=[(an+1)+(a

Answer 1

(1^2+...+n^2=n(n+1)(2n+1)/6 这是基本结论)

由(7n+1)/6=[a(n+1)+...+a(2n)]/S(n)

S(n)=[6/(7n+1)][a(n+1)+...+a(2n)]

S(n)=[6/(7n+1)]{[1^2+...+(2n)^2]-(1^2+...+n^2)}

S(n)=[6/(7n+1)][(2n)(2n+1)(4n+1)/6-n(n+1)(2n+1)/6]

S(n)=[1/(7n+1)](7n^2+n)(2n+1)

S(n)=2n^2+n

∴b(n)=S(n)-S(n-1)=4n-1(n＞1)

b(1)=S(1)=3满足4n-1

∴b(n)=S(n)-S(n-1)=4n-1

(2)证明：

2^[-b(n)]/2^[-b(n-1)]

=2^[b(n-1)-b(n)]

=2^(-4)

=1/16

∴数列{2^[-b(n)]}是等比数列