I'm trying to get the simple form of sum{k=1 to n} _2k C _k... I've tried to use Pascal's triangle, the gamma function, etc... but I couldn't get it... Somebody help~ My guess is _2k C _k is equal to binomial(2k,k). I used Maple... and the answer is ... as I see, a complex number!

F:=binomial(2*k,k);
                              binomial(2 k, k)
DefiniteSummation(F,k=1..n);

         (1/2)                                                               
    4 I 3                                            /[       3]            \
- -------------- - binomial(2 n + 2, n + 1) hypergeom|[1, n + -], [n + 2], 4|

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