Question

在数列{an}中,a1=1,2an+1=(1+1/n)^2*an，证明：数列{an/n^2}是等比数列，并求an的通项公式

在数列{an}中,a1=1,2an+1=(1+1/n)^2*an，证明：数列{an/n^2}是等比数列，并求an的通项公式;(2)令bn=an+1-1/2an,求数列{bn}的前n相和

Answer 1

1,2a(n+1)=(1+1/n)^2*an 2a(n+1)=[(n+1)/n]^2*an a(n+1)/(n+1)^2=(1/2)(an/n^2)
所以，数列{an/n^2}是首项为1、公比为1/2的等比数列，an/n^2=(1/2)^(n-1)
an=n^2*(1/2)^(n-1)(n1,2,3,……,)
2,bn=(n+1)^2*(1/2)^n-n^2*(1/2)^n=(2n+1)(1/2)^n。设Tn=b1+b2+…+bn，则
Tn=3*(1/2)+5*(1/2)^2+7*(1/2)^3+…+(2n-1)*(1/2)^(n-1)+(2n+1)*(1/2)^n (1)
(1/2)*(1)得：
(1/2)Tn=3*(1/2)^2+5*(1/2)^3+7*(1/2)^4+…+(2n-1)*(1/2)^n+(2n+1)*(1/2)^(n+1) (2)
(1)-(2)得：
(1/2)Tn=1/2+2*(1/2)+2*(1/2)^2+2*(1/2)^3+…+2*(1/2)^n-(2n+1)*(1/2)^(n+1)
=1/2+2*(1/2)*[1-(1/2)^n]/(1-1/2)-(2n+1)*(1/2)^(n+1)
=1/2+2-(1/2)^(n-1)-(2n+1)*(1/2)^(n+1)
=3/2-(1/2)^(n-1)-(2n+1)*(1/2)^(n+1)
Tn=3-(1/2)^(n-2)-(2n+1)*(1/2)^n

Answer 2

题目不清