- #1
pig
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I tried induction, but got no results. What I get is:
f(n+1) = 4f(n) - 6f(n)/(n+2)
So now I'd have to prove that 6*(2n)!/[n!(n+1)!(n+2)] is natural, which would be going in circles.
I tried to go directly, but came up with nothing good. It can be written as [2n over (n-1)] / n, but I still didn't manage to prove it.
Can anyone help?
By the way, I got this problem from a high school kid who was supposed to prove it by induction, so I presume I'm missing something obvious here?
Thanks.
f(n+1) = 4f(n) - 6f(n)/(n+2)
So now I'd have to prove that 6*(2n)!/[n!(n+1)!(n+2)] is natural, which would be going in circles.
I tried to go directly, but came up with nothing good. It can be written as [2n over (n-1)] / n, but I still didn't manage to prove it.
Can anyone help?
By the way, I got this problem from a high school kid who was supposed to prove it by induction, so I presume I'm missing something obvious here?
Thanks.