题目

数列An的前n项和为Sn,并且Sn等于n²-4n,设Bn=An÷（2的n次幂）,求数列Bn的前n项和

提问时间：2021-03-13

答案

因为Sn=n^2-4n,Sn+1=（n+1）^2-4（n+1）,两式相减,可得a(n+1)=2n-3,所以an=2n-5
bn=an/2^n=(2n-5)/2^n
Tn=-3/2-1/2^2+1/2^3+3/2^4+.+(2n-5)/2^n
2Tn=-3-1/2+1/2^2+3/2^3+.+(2n-5)/2^(n-1)
2Tn-Tn=-3+2(1/2+1/2^2+1/2^3+...+1/2^(n-1))-(2n-5)/2^n
Tn=-3+2*1/2*(1-1/2^(n-1))/(1-1/2)-(2n-5)/2^n
=-3+2-4/2^n-(2n-5)/2^n
=-1-(2n-1)/2^n