Question

已知数列an是首项和为1,公比为2的等比数列,bn的前n项和sn=n^2

Answer 1

1、an = 2^(n-1),b1=S1=1,bn = Sn - Sn-1 = n^2 - (n-1)^2 = 2n-1,n>=2,n=1也合适 bn = 2n-1

2、令 cn= bn/an = 2n-1 / 2^(n-1) 错位相减:

Sn = c1+c2+...+cn = 1/1 + 3/2 + 5/4 +...+ 2n-3 / 2^(n-2) + 2n-1 / 2^(n-1)

1/2*Sn = 1/2 + 3/4 +...+ 2n-5 / 2^(n-2) + 2n-3 / 2^(n-1) + 2n-1 / 2^n

相减得 Sn - 1/2*Sn = 1/1 + { 2/2 + 2/4 +...+ 2 / 2^(n-2) + 2 / 2^(n-1) } - 2n-1 / 2^n

1/2*Sn = 1 + [ 1 + 1/2 + ...+ 1/2^(n-2) ] - 2n-1 / 2^n

= 1 + 2 - 1/2^(n-2) - 2n-1 / 2^n = 3 - 2n+3 / 2^n

Sn = 6 - 2n+3 / 2^(n-1)