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Posted by longpo at 2009-05-19 05:22:32 on Problem 3734
In Reply To:Re:递推公式 a[i]=2*(a[i-1]+a[i-2]*(i-1)) ? Posted by:demo at 2009-05-18 21:21:57

>     n/2            (n-2*k)/2
> ans=sigma C(n,2*k)* sigma C(n-2*k,2*m)*2^(n-2*k-2*m)
>     k=0             m=0
>     n/2
>    =sigma C(n,2*k)*2^(n-2*k)*1/2*[(1/2)^(n-2*k)+(3/2)^(n-2*k)]
>     k=0
>         n/2
>    =1/2*sigma C(n,2*k)*[1+3^(n-2*k)] 
>         k=0
>         n/2                    n/2 
>    =1/2*sigma C(n,2*k)+1/2*3^n*sigma C(n,2*k)*3^(-2*k)
>         k=0                    k=0
>    =2^(n-1)+4*(n-1)
> 其中化简用到二项式展开

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