一道数列题：在数列{an}中 ,a1=1,2A（n+1）=(1+1/n)2an.求通项公式.
3个回答
(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)
相关问题
