Question

计算数列{(1-1/4)(1-1/9)(1-1/16)...[1-1/(n+1)^2]}的前4项的值

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)]