AL pure maths問題,極急!!

(ii)

I(n+2)

= (n+1)/(n+2)I(n)

= [(n+1)/(n+2)][(n-1)/n]I(n-2)

= ...

= [(n+1)/(n+2)][(n-1)/n]....(1/2)I(0)

Since I(0) = π^2/2, the result follows.

2011-04-04 21:09:59 補充：
Some typing mistake

I(2n+2)

= (2n+1)/(2n+2)I(2n)

= [(2n+1)/(2n+2)][(2n-1)/2n]I(2n-2)

= ...

= [(2n+1)/(2n+2)][(2n-1)/2n]....(1/2)I(0)