Question

设数列{an}的前n项和Sn=2an-2^n

1.求a3，a4；
2.证明：{an+1-2an}是等比数列
3.求{an}的通项公式
【数列问题】
希望各位慷慨解囊，有详细过程
谢谢~o(≥v≤)o~~

Answer 1

1.
A1=S1=2A1-2^1 A1=2
S2=A1+A2=2A2-2^2 A2=6
S3=S2+A3=2A3-2^3 A3=16
S4=S3+A4=2A4-2^4 A4=40

2.
Sn=2An-2^n
S(n+1)=2A(n+1)-2^(n+1)
两式相减
A(n+1)=2A(n+1)-2An-2×2^n+2^n
A(n+1)-2An=2^n
A2-2A1=6-2×2=2
{A(n+1)-2An}是等比数列

3.
An-2A(n-1)=2^(n-1)
An/2^n-A(n-1)/2^(n-1)=1/2
A(n-1)/2^(n-1)-A(n-2)/2^(n-2)=1/2
……
A2/2^2-A1/2^1=1/2
上式相加，相同项消去。
An/2^n-A1/2=(n-1)/2
An=(n+1)/2×2^n=(n+1)×2^(n-1)

Answer 2

1.
S1 = a1 = 2 a1 - 2 , => a1 = 2
S2 = a1 + a2 = 2 + a2 = 2 a2 - 4 , => a2 = 6
S(n+1) - Sn = a(n+1) = [2 a(n+1) - 2^(n+1)] - (2 an - 2^n)
= 2 a(n+1) - 2 an - 2 * 2^n + 2^n
= 2 a(n+1) - 2 an - 2^n
所以 a(n+1) - 2 an = 2^n
a3 = 2 a2 + 4 = 16 , a4 = 2 a3 + 8 = 40 .
2.
因为 a(n+1) - 2 an = 2^n , a2 - 2 a1 = 2 ,
所以 {an+1 - 2 an}是等比数列,公比为 2 ，首项为 2 。
3.
an = 2 a(n-1) + 2^(n-1)
= 2 [2 a(n-2) + 2^(n-2)] + 2^(n-1)
= 2^2 a(n-2) + 2^(n-1) + 2^(n-1)
= 2^2 [2 a(n-3) + 2^(n-3)] + 2^(n-1) + 2^(n-1)
= 2^3 a(n-3) + 2^(n-1) + 2^(n-1) + 2^(n-1)
= …
= 2^(n-1) a1 + (n-1) 2^(n-1)
= (n + 2) 2^(n - 1)