Question

数列an中,a1=1/2,an=(n-1)/(n+1)乘以an-1,求an的通项.

Answer 1

n>=2时

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

=[(n-1)/(n+1)]*[ (n-2)/n] a(n-2)

=[(n-1)/(n+1)]*[(n-2)/n ]*[(n-3)/(n-1) ]a(n-2)

=[(n-1)/(n+1)]*[(n-2)/n ]*[(n-3)/(n-1) ]*.*[(4-1)/(4+1)][(3-1)/(3+1)][(2-1)/(2+1)]a1

=(3-1)(2-1)/[(n+1)(n)]*(1/2)

=1/[n(n+1)]

n=1时 a1=1/2=1/[(1)(1+1)]

所以对正自然数n

an=1/[n(n+1)]