求数列{(n+1)⼀2的n次幂}的前n项和

Sn= 2/2+3/2^2+4/2^3+5/2^4+……+ (n-1)/2^(n-2)+n/2^(n-1)+(n+1)/2^n (1)
2Sn=2+3/2+4/2^2+5/2^3+……+(n-1)/2^(n-3)+n/2^(n-2)+(n+1)/2^(n-1) (2)
(2)－(1)得
Sn＝2+1/2+1/2^2+1/2^3+……+1/2^(n-1)－(n+1)/2^n
＝1+[1+1/2+1/2^2+1/2^3+……+1/2^(n-1)]－(n+1)/2^n
＝1+(1-1/2^n)/(1-1/2)－(n+1)/2^n
＝3－[(n+3)/2^n]

an=(n+1)/2ⁿ=n/2ⁿ +1/2ⁿ
Sn=a1+a2+...+an=(1/2+2/2²+...+n/2ⁿ)+(1/2+1/2²+...+1/2ⁿ)
令An=1/2+2/2²+...+n/2ⁿ
则An/2=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
An -An/2=An/2=1/2+1/2²+...+1/2ⁿ -n/2^(n+1)
=(1/2)(1-1/2ⁿ)/(1-1/2) -n/2^(n+1)
=1-1/2ⁿ -n/2^(n+1)
An=2-2/2ⁿ-n/2ⁿ
Sn=An+(1/2+1/2²+...+1/2ⁿ)
=2-2/2ⁿ -n/2ⁿ +(1/2)(1-1/2ⁿ)/(1-1/2)
=2-2/2ⁿ-n/2ⁿ+1-1/2ⁿ
=3-(n+3)/2ⁿ