$\sum_{l=0}^{k/2} \frac{k!}{(k-2l)!l!}$
I want to calculate the above sum. Its form is similar to that of binomial but not same.
How to calculate it? Am I in wrong path?(which means its solution is not relevant to binomial?)
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$$1=1\\ 1+1=2\\ 1+6+1=8\\ 1+10+5=16\\ 1+15+15+1=32$$
isn't this enough ?
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I
m sorry but I cant understand how that numbers are related to the solution. Could you give me little bit more? – kyle Apr 24 '17 at 08:13 -
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I totally can
t understand what your intention is. I know that is pascals triangle. And I know it is related to binomial coefficient. But I cant understand how that is related to my question. I tried as best as I can to change it into the form that fits it but I cant. Thats why I questioned it here. I also cant understand how the numbers you show is related to my question. I calculate the sum of my question when k = 0 , 1 ,2 ,3, 4 ... and it doesn`t fit yours at all. – kyle Apr 24 '17 at 09:25 -
ah... hey, are you thinking that my question is $\sum_{l=0}^{k/2} \frac{k!}{(k-2l)!(2l)!}$ ? – kyle Apr 24 '17 at 09:33
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