计算数列{(1-1/4)(1-1/9)(1-1/16)····[1-1/(n+1)^]}的通项公式..

Question

如题所述

Answer 1

a1=(1-1/4)=3/4
a2=(1-1/4)(1-1/9)
=3/4*8/9
=1/3=4/12
a3=a2*(1-1/16)=1/3*15/16=5/16
a4=a3*(1-1/25)=5/16*24/25=3/10=6/20
所以通项从第二项开始猜测为分母=n+2
分子=4(n+1)
所以an=(n+2)/[4(n+1)],n>=2
a1=3/4
对于n=2显然成立
假设对于n=k成立
即ak=(k+2)/[4(k+1)]
n=k+1时
a(k+1)=ak*(1-1/(k+2)^2)
=(k+2)/[4(k+1)]*[(k+2)^2-1]/(k+2)^2
=(k+2)/[4(k+1)]*[(k+3)(k+1)]/(k+2)^2
=(k+3)/4(k+2)
=[(k+1)+2]/[4[(k+1)+1]]
所以对于任意自然数n>=2都有an=(n+2)/[4(n+1)]追问

不要复制黏贴过来..你算算3/4*8/9到底等于多少...

Answer 2

an=1-1/(n+1)(n+1)