Question

求数列2分之1,4分之2,8分之3,16分之4……的前n项和

Answer 1

通项an=n/(2^n)=n×2^(-n)

∴ Sn=1×2^(-1)+2×2^(-2)+3×2^(-3)+……+n×2^(-n) ①

∴2Sn=1×2^0+2×2^(-1)+3×2^(-2)+……+n×2^(1-n) ②

②-①得 Sn=[2^0+2^(-1)+2^(-2)+……+2^(1-n)]-n×2^(-n)

=2-2^(1-n)-n×2^(-n)