题目

cn二1÷(4n^2一1),求证前n项和tn

提问时间：2021-04-27

答案

bn=2^(3-2n)即2的3-2n次方
cn=an/bn=(2n-1)*2^(2n-2)
Tn=c1+c2+……+cn=1*2^0+3*2^2+5*2^4+……+(2n-1)*2^(2n-2)
4Tn=1*2^2+3*2^4+……+2(n-2)*2^(2n-2)+2(n-1)*2^2n
4Tn-Tn=3Tn=2(n-1)*2^2n-2*[2^2+2^4+……+2^(2n-2)]-1*2^0
=2(n-1)*2^2n-2^3[1-4^(n-1)] / (1-4) -1
=(n-1)*2^(2n+1)-[2^(2n+1)-8] / 3 -1
Tn=5/9+(9n-12)*2^(2n+1)/9