Question

若数列bn=2n除以2的n次方 求前n项和Sn

Answer 1

n=2n/2^n

Sn=b1+b2+b3+.+b(n-1)+bn

=2·1/2+2x2·1/2²+2x3·1/2³+.+2n·1/2^n .①

1/2·Sn= 2·1/2²+2x2·1/2³+.+2(n-1)·1/2^n +2n·1/2^(n+1).②

①-②得：1/2·Sn=2(1/2+1/2²+1/2³+.+1/2^n)-2n·1/2^(n+1)

=2·1/2（1-1/2^n）/(1-1/2)-2n·1/2^(n+1)

∴sn=[2^(n+1)-n-2]/2^(n-1)