Question

已知数列{an}中,a1=5,an=2an-1=2^n-1(n∈N*且大于等于2),求数列{an}的前n项和Sn

Answer 1

是“an=2a(n-1)+2^n-1”吧？

an=2a(n-1)+2^n-1
an - 1 = 2×[ a(n-1) - 1 ] +2^n， 两边同时除以2^n，得
(an - 1)/2^n = 2×[ a(n-1) - 1 ]/2^n +1
(an - 1)/2^n = [ a(n-1) - 1 ]/2^(n-1) +1
(an - 1)/2^n - [ a(n-1) - 1 ]/2^(n-1) =1，n≥2
即数列{(an - 1)/2^n}是以(a1-1)/2=2为首项，1为公差的等差数列
∴(an - 1)/2^n=2+(n-1)×1=n+1
an - 1=(n+1)·2^n
令数列{an - 1}的前n项和为Tn
则 Tn= 2×2 + 3×2² + 4×2³ + …… + (n+1)·2^n .........................①
2Tn= 2×2² + 3×2³ + …… + n·2^n + (n+1)·2^(n+1) ..........................②
②－①，得
Tn= (n+1)·2^(n+1) - (2²+2³+……+2^n) - 4
= (n+1)·2^(n+1) - (2+2²+2³+……+2^n) - 2
= (n+1)·2^(n+1) - [2(1-2^n)/(1-2)] - 2
= (n+1)·2^(n+1) - 2^(n+1)
= n·2^(n+1)
∴Sn=Tn + n=n·2^(n+1) + n

Answer 2

题目麻烦表达清楚一点……

Answer 3

Sn=1（n=1）Sn=2^(n-1)-1（n>=时）