Question

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

Answer 1

Sn=n²-4n
=>a1=S1=-3
an=Sn-S(n-1)=2n-5
=>{an}是以-3为首项，-5为公差的等差数列。
=>bn=(2n-5)÷2^n
=>bn的前n项和
Bn=a1/2+a2/2^2+a3/2^3+....+an/2^n..........(1)
2Bn=a1+a2/2+a3/2^2+....+an/2^(n-1)..........(2)
(2)-(1)得
Bn=a1-5(1/2+1/2^2+...+1/2^(n-1)=-3-5[(1/2)(1-(1/2)^(n-1))]/[1-1/2)]=5/2^(n-1)]-8

Answer 2

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