Question

一道数列题：在数列{an}中 ，a1=1,2A（n+1）=(1+1/n)2an。求通项公式。

请问下这题数列,a1=1,2A（n+1）=(1+1/n)2an。 一。求通项公式。二。令bn=a（n+1）-1/2an，求bn前n项数列和Sn。三。求数列{an}前n项和Tn

Answer 1

(1)2a(n+1）=(1+1/n)² an
a(n+1)/an=1/2 *(n+1)²/n²
n≥2
a2/a1=4/1 *1/2
a3/a2=9/4*1/2
a4/a3=16/9*1/2
...................
an/a(n-1)=n^2/(n-1)^2 *1/2
an/a1=n^2*(1/2)^(n-1)
an=n^2*(1/2)^(n-1)
n=1时，上式成立
所以，an=n^2*(1/2)^(n-1) (n∈N*)
(2)bn=(n+1)^2/2^n -1/2 *n^2*(1/2)^(n-1)=(2n+1)/2^n
Sn=3/2+5/4+7/8+......+(2n+1)/2^n 1)
1/2Sn=3/4+5/8+........+(2n-1)/2^n+(2n+1)/2^(n+1) 2)
1)-2):
1/2Sn=3/2+2/4+2/8+......+2/2^n-(2n+1)/2^(n+1)
=3/2+1/2[1-1/2^(n-1)]/(1-1/2)-(2n+1)/2^(n+1)
=5/2-1/2^(n-1)-(2n+1)/2^(n+1)
=5/2-(2n+5)/2^(n+1)
Sn=5-(2n+5)/2^n
(3)a(n+1)=(n+1)^2*(1/2)^n
∵bn=a（n+1）-1/2an
两边求和左边为Sn，右边为a2+a3+...+a(n+1)-1/2Tn
Sn=T(n+1)-1/2 Tn==>Sn=Tn-a1+a(n+1)-1/2Tn
==>Tn=2Sn+2-2a(n+1)=12-(2n+5)/2^(n-1)-2(n+1)^2*(1/2)^n=12-(n^2+4n+6)/2^(n-1)

Answer 2

a1=1
a2=(1+1/1)=2
a3=(1+1/2)2=3
a4=(1+1/3)3=4
……
an=n
a(n+1)=(1+1/n)n=n+1

bn=n+1-n/2=n/2+1
Sn=(1+n)n/4+n

Tn=(1+n)n/2

Answer 3

1,令√an=Cn
则C(n+1)/Cn=(n+1)/√2n
所以Cn/C1=Cn/C(n-1)*C(n-1)/C(n-2)*...*C2/C1=(√2/2)^(n-1)*n
所以an=(1/2)^(n-1)*n^2
2,bn=(1/2)^n*(n+1)^2-(1/2)^(n+1)*n^2=(1/2)^n(2n+1)
所以bn前n项和
Sn=(1/2)*3+(1/2)^2*5+....+(1/2)^n(2n+1) ①
Sn/2=(1/2)^2*3+(1/2)^3*5+...+(1/2)^(n+1)(2n+1) ②
两式相减得
Sn/2=3/2+2[(1/2)^2+(1/2)^3+...+(1/2)^n]-(1/2)^(n+1)(2n+1)
=3/2+1-(1/2)^(n-1)-(1/2)^(n+1)(2n-1)
Sn=5-(1/2)^n(2n+5)
3,Tn=1^2*(1/2)^0+2^2*(1/2)^1+...+n^2*(1/2)^(n-1) ③
Tn/2=1^2*(1/2)^1+...+(n-1)^2*(1/2)^(n-1)+n^2*(1/2)^n ④
两式相减得
Tn/2=1+[3*(1/2)^1+5*(1/2)^2+...+(2n-1)*(1/2)^(n-1)]-n^2*(1/2)^n
令Tn/2+n^2*(1/2)^n-1=kn ⑤
则
kn=3*(1/2)^1+5*(1/2)^2+...+(2n-1)*(1/2)^(n-1)
比较①式容易看出kn=S(n-1)
把Sn带入⑤式解得
Tn=12-(1/2)^(n-1)(n^2+4n+6)
PS：凌晨两点，头有点晕，不知道自己答案算对没有,不过可以看出1楼看不懂，2楼3问计算有点错。

不懂再问，For the lich king