在数列{an}中,a1=1,2a(n+1)=(1+1/n)^2*an,（1）证明数列{an/n^2}是等比数列,并求{a

在数列{an}中,a1=1,2a(n+1)=(1+1/n)^2*an,（1）证明数列{an/n^2}是等比数列,并求{an}的通项公式
（2）令bn=a(n+1)-1/2an,求数列{bn}的前n项和Sn;(3)求数列{an}的前n项和Tn

1.
2a(n+1)=(1+ 1/n)²an=[(n+1)²/n²]an
2a(n+1)/(n+1)²=an/n²
[a(n+1)/(n+1)²]/(an/n²)=1/2,为定值.
a1/1²=1/1=1
数列{an/n²}是以1为首项,1/2为公比的等比数列.
an/n²=1×(1/2)^(n-1)=1/2^(n-1)
an=n²/2^(n-1)
n=1时,a1=1²/2^0=1/1=1,同样满足通项公式
综上得数列{an}的通项公式为an=n²/2^(n-1)
2.
bn=a(n+1) -(1/2)an
=(n+1)²/2ⁿ -(1/2)[n²/2^(n-1)]
=(n+1)²/2ⁿ -n²/2ⁿ
=[(n+1)²-n²]/2ⁿ
=(2n+1)/2ⁿ
=n/2^(n-1) +1/2ⁿ
Sn=b1+b2+...+bn=1/1+2/2+3/2²+...+n/2^(n-1) +(1/2+1/2²+...+1/2ⁿ)
令Cn=1/1+2/2+3/2²+...+n/2^(n-1)
则Cn/2=1/2+2/2²+...+(n-1)/2^(n-1)+n/2ⁿ
Cn-Cn/2=Cn/2=1+1/2+1/2²+...+1/2^(n-1) -n/2ⁿ
=1×(1-1/2ⁿ)/(1-1/2) -n/2ⁿ
=2-(n+2)/2ⁿ
Sn=Cn+(1/2+1/2²+...+1/2ⁿ)
=2-(n+2)/2ⁿ +(1/2)(1-1/2ⁿ)/(1-1/2)
=2-(n+2)/2ⁿ+1-1/2ⁿ
=3- (n+3)/2ⁿ
3.
an=n²/2^(n-1)
Tn=a1+a2+...+an=1/1+2²/2+3²/2²+...+n²/2^(n-1)
Tn/2=1/2+2²/2²+...+(n-1)²/2^(n-1)+n²/2ⁿ
Tn-Tn/2=Tn/2=1+(2²-1²)/2+(3²-2²)/2²+...+[n²-(n-1)²]/2^(n-1) -n²/2ⁿ
=[1/2^0+3/2+5/2²+...+(2n-1)/2^(n-1)]-[1/2+1/2²+...+1/2^(n-1)]-n²/2ⁿ
令Dn=1/2^0+3/2+5/2²+...+(2n-1)/2^(n-1)
则Dn/2=1/2+3/2²+...+(2n-3)/2^(n-1)+(2n-1)/2ⁿ
Dn-Dn/2=Dn/2=1+2/2+2/2²+...+2/2^(n-1)-(2n-1)/2ⁿ
=2[1+1/2+1/2²+...+1/2^(n-1)] -1 -(2n-1)/2ⁿ
=2×1×(1-1/2ⁿ)/(1-1/2) -1-(2n-1)/2ⁿ
=3-(2n+3)/2ⁿ
Dn=6-(4n+6)/2ⁿ
Tn/2=Dn -[1/2+1/2²+...+1/2^(n-1)]-n²/2ⁿ
=6-(4n+6)/2ⁿ-(1/2)[1-1/2^(n-1)]/(1-1/2) -n²/2ⁿ
=5- (n²+4n+4)/2ⁿ
Tn=10- 2(n²+4n+4)/2ⁿ=10 -(n+2)²/2^(n-1)
提示：第3问比较麻烦,需要两次运用错位相减法.