Question

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

Answer 1

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。

Answer 2

(1)2a(n+1)=(1+1/n)^2*an=[(n+1)^2/n^2]a(n) a(n+1)/(n+1)^2=(1/2) a(n)/n^2
a(n)/n^2 为公比为1/2的等比数列
a(n)/n^2=(1/2)^(n-1) a(1)/1^2=1/2^(n-1)
a(n)=n^2/2^(n-1)
(2) b(n)=a(n+1)-(1/2)a(n)=(n+1)^2/2^n-n^2/2^n=(2n+1)/2^n
Sn=3/2+5/2^2+..................+(2n-1)/2^(n-1)+(2n+1)/2^n
2Sn=3+ 5/2+......................+(2n-1)/2^(n-2)+(2n+1)/2^(n-1)
Sn=2Sn-Sn=3+2/2+2/2^2+....................+2/2^(n-1)-(2n+1)/2^n=5-1/2^(n-2)-.(2n+1)/2^n
=5-(2n+5)/2^n
(3) 用求Sn类似的方法可求出Tn=12-2(n²+4n+6)/2^n

Answer 3

1. 将2a(n+1)=(1+1/n)^2*an式子化成2(an+1)/(n+1)²=an/n²,又因为根据2a(n+1)=(1+1/n)^2*an，得出a2=2，所以an/n^2是个首项是1，公比为2的等比数列。

2. 因为an/n^2=2的n次方，所以an=n^2×2∧n