在数列{an}中 a1=1 2a(n+1)=(1+1/n)^2乘an 【a(n+1)是的第n+1项】

在数列{an}中 a1=1 2a(n+1)=(1+1/n)^2乘an 【a(n+1)是的第n+1项】
(1)求an
（2）另bn=a(n+1)-0.5an 求数列bn的前n项和Sn
（3）求数列an的前n项和Tn

LS未写出第(3)问
注意:(2)用错位相减法,此处将过程略去.
(1)
2a(n+1)=(1+1/n)^2*an
则:
a(n+1)/an=(1/2)*[(n+1)/n]^2
则有:
an/a(n-1)=(1/2)*[n/(n-1)]^2
a(n-1)/a(n-2)=(1/2)*[(n-1)/(n-2)]^2
...
a2/a1=(1/2)*[2/1]^2
将上式累乘,得:
an/a1
=(1/2)^(n-1)*[n/(n-1)*(n-1)/(n-2)*...*2/1]^2
=(1/2)^(n-1)*n^2
由于a1=1
则:
an=n^2*(1/2)^(n-1)
(2)
bn=a(n+1)-0.5an
=(n+1)^2*(1/2)^n-(1/2)*[n^2*(1/2)^(n-1)]
=(2n+1)*(1/2)^n
故Sn=b1+b2+...+bn
=3*(1/2)^1+5*(1/2)^2+...+(2n+1)*(1/2)^n
=5-(1/2)^(n-2)-(2n+1)*(1/2)^n
=5-(2n+5)*(1/2)^n
(3)
Tn=a1+a2+...+an
=1^2*(1/2)^0+2^2*(1/2)^1+...+n^2*(1/2)^(n-1) -----(1)
则:
(1/2)Tn=1^2*(1/2)^1+...+(n-1)^2*(1/2)^(n-1)+n^2*(1/2)^n -----(2)
(1)-(2)得:
(1/2)Tn=1+[3*(1/2)^1+5*(1/2)^2+...+(2n-1)*(1/2)^(n-1)]-n^2*(1/2)^n
(1/2)Tn=1+[Sn-(2n+1)*(1/2)^n]-n^2*(1/2)^n
=1+Sn-(n+1)^2*(1/2)^n
=6-(n^2+4n+6)*(1/2)^n
则:Tn=12-(n^2+4n+6)*(1/2)^(n-1)