Question

数列an=n,bn=1/2的n次方,求a1b1+a2b2+...anbn的值

Answer 1

Sn=a1b1+a2b2+...anbn=1*(1/2)+2(1/4)+3(1/8)+4(1/16)+.n(1/2^n)①

1/2Sn=1*(1/4)+2(1/8)+3(1/16)+4(1/32)+.n(1/2^（n+1）)②

①-②得

1／2Sn=1／2＋1／4＋1／8＋1／16＋...＋1／2^n－n(1/2^(n+1))=[2^(n+1)－n－2]／（2^(n+1))

Sn=[2^(n+1)－n－2]／2^n