已知数列分子为n锧1,分母为2的n次方,An=(n锧1)⼀2^n,求前n项和Sn

2024-11-30 22:59:14
推荐回答(1个)
回答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