求数列二分之二,二的平方分之四,二的三次方分之六,…,二的n次方分之2n,的前n项和

2025-12-15 05:09:35
推荐回答(1个)
回答1:

Sn=2/2+ 4/2²+ 6/2³+......+2n/2^n
(Sn)/2= 2/2²+4/2³+.......+(2n-2)/2^n+2n/2^(n+1)
两式对应相减得:(Sn)/2=2/2+2/2²+2/2³+........+2/2^n-2n/2^(n+1)
=(1-1/2^n)/(1-1/2)-2n/2^(n+1)
=2-1/2^(n-1)-n/2^n
得:Sn=4-1/2^(n-2)-n/2^(n-1)