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Sn;(3)求数列{an}的前n相和Tn

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;(2)令bn=an+1-1/2an,求数列{bn}的前n相和Sn;(3)求数列{an}的前n相和Tn

Answer 1

(1)
2a(n+1)=(1+1/n)^2*an
= [(n+1)/n]^2 .an
2a(n+1)/(n+1)^2 = an/n^2
[a(n+1)/(n+1)^2]/[an/n^2] = 1/2
=>{an/n^2}是等比数列
[a(n+1)/(n+1)^2]/[an/n^2] = 1/2
[an/n^2]/[a1/1^2]=(1/2)^(n-1)
an/n^2 = (1/2)^(n-1)
an = n^2 .(1/2)^(n-1)
(2)
bn = a(n+1) -(1/2)an
= (n+1)^2.(1/2)^n - (1/2)[n^2 .(1/2)^(n-1)]
= (1/2)^n .(2n+1)
= n.(1/2)^(n-1) + (1/2)^n
consider
1+x+x^2+..+x^n = (x^(n+1)-1)/(x-1)
1+2x+..+nx^(n-1)=[(x^(n+1)-1)/(x-1)]'
= [ nx^(n+1)-(n+1)x^n +1] /(x-1)^2
put x=1/2
1.(1/2)^0+ 2.(1/2)^1 +...n(1/2)^(n-1)
=4[ n.(1/2)^(n+1)-(n+1).(1/2)^n +1]
Sn =b1+b2+..+bn
= 4[ n.(1/2)^(n+1)-(n+1).(1/2)^n +1] + ( 1-(1/2)^n )
= (1/2)^n .( 2n - 4(n+1) -1 ) + 5
= 5 - (1/2)^n .(2n+5)