数列（2n+1）（2分之1）的n次方的前n项和

利用错位相减法即可求解。
令an=(2n+1)/(2^n)
Sn是{an}的前n项和
则Sn=3/2+5/4+7/8+……+(2n+1)/(2^n) ①
乘2
2Sn=3+5/2+7/4+……+(2n+1)/(2^(n-1)) ②
错位相减
得Sn= 5 -(2n+5)/(2^n)

an =(2n+1)(1/2)^n
= n(1/2)^(n-1) + (1/2)^n
Sn =a1+a2+..+an
= {summaiton(i:1->n)i.(1/2)^(i-1)} + (1-(1/2)^n)
consider
1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1)
1+2x+..+nx^(n-1) =[(x^(n+1)- 1)/(x-1)]'
= [nx^(n+1) - (n+1)x^n + 1]/(x-1)^2
put x=1/2
summation(i:1->n) i.(1/2)^(i-1)
= 4(n.(1/2)^(n+1) - (n+1).(1/2)^n + 1)
= 4[1- (n+2).(1/2)^(n+1)]

Sn = {summaiton(i:1->n)i.(1/2)^(i-1)} + (1-(1/2)^n)
= 4[1- (n+2).(1/2)^(n+1)] +(1-(1/2)^n)
= 5 -(2n+5)(1/2)^n