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

问题描述:

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

Sn=a1b1+a2b2+...anbn=1*(1/2)+2(1/4)+3(1/8)+4(1/16)+.n(1/2^n)①1/2Sn=1*(1/4)+2(1/8)+3(1/16)+4(1/32)+.n(1/2^(n+1))②①-②得1/2Sn=1/2+1/4+1/8+1/16+...+1/2^n-n(1/2^(n+1))=[2^(n+1)-n-2]/(...