Question

已知数列分子为n➕1,分母为2的n次方,An=（n➕1）/2^n,求前n项和Sn

Answer 1

An=（n➕1）/2^n

Sn=2/2+3/2^2+4/2^3+.+(n+1)/2^n ①

①×1/2:

1/2Sn=2/2^2+3/2^3+4/2^4+.+n/2^n+(n+1)/2^(n+1) ②

①-②:

1/2Sn=1+1/2^2+1/2^3+...+1/2^n-(n+1)/2^(n+1)

=[1-1/2^n]/(1-1/2)-(n+1)/2^(n+1)

=2-4/2^(n+1)-(n+1)/2^(n+1)

=2-(n+5)/2^(n+1)

Sn=4-(n+5)/2^n