Question

已知数列{an},an=(n+1)*2^n,求Sn

Answer 1

Sn=a1+a2+a3+...+a(n-1)+an=(1+1)*2^1+(1+2)*2^2+(1+3)*2^3+...+(1+(n-1))*2^(n-1)+(1+n)*2^n
Sn=2*2+3*2^2+4*2^3+...+n*2^(n-1)+(n+1)*2^n (1)
2Sn=2*2^2+3*2^3+4*2^4+...+n*2^n+(n+1)*2^(n+1) (2)
(2)-(1)得
Sn=2Sn-Sn=(n+1)*2^(n+1)-(2^2+2^3+...+2^n)-2*2=(n+1)*2^(n+1)-2^(n+1)-8=n*2^(n+1)-8
所以Sn=n*2^(n+1)-8

Answer 2

a1=2*2=4
a(n+1)=(n+2)*2(n+1)=2(n+1+1)*2^n=2*(n+1)*2^n+2*2^n=2an+2^(n+1)
an=2a(n-1)+2^n
a(n+1）=2（2a(n-1)+2^n)+2^(n+1)=2*2a(n-1)+2^(n+1)+2^(n+1)
=2*2a(n-1)+2*2^(n+1)
so a(n+1)=2^n*a1+2^n*2^(n+1)=2^n*4+2*4^n
Sn=4(2+4+8+...+2^(n-1))+2(4+16+....+4^(n-1))
=4*2*（2^(n-1)-1)+2*(4*(1-4^(n-1))/(1-3)
=2^(n+2)-8+4^n-4=2^(n+2)+4^n-12

Answer 3

Sn=n*2^(n＋1)
利用Sn两边同乘2可解

Answer 4

Answer 5

错位相减法可求解。

追问

求过程—A—

追答

Sn....1
(1/2)Sn.....2
1-2
=>.....