Calculating $\sum_{k=0}^{13} \binom{26}{2k}\frac{(2k)!}{k!2^k}$

Question

it may be an odd request but I'm trying to calculate a certain number and atm I lack any advance software like MATLAB or Maple. I also failed to find any good calculator online that would be able to calculate it. I want to use it for further calculations regarding permutations.

Can anyone calculate it for me?

$\sum_{k=0}^{13} \binom{26}{2k}\frac{(2k)!}{k!2^k}$

Thanks!

You could also use a spreadsheet. – NoChance Oct 03 '19 at 10:32 — NoChance, Oct 03 '19 at 10:32

Toby Mak · Accepted Answer · 2019-10-03T10:34:05.443

0

I get $532 \ 985 \ 208 \ 200 \ 576$ using Wolfram Alpha. This number is approximately $5.33 \times 10^{14}$.

For your particular question, it helps a lot to note that $\sum_{\text{i even}} {n \choose i} = 2^{n-1}$ (this post). In your case then, you have $\sum_{k=0}^{13} {26 \choose {2k}} = 2^{25}$. Most modern calculators (not graphing calculators) have a button for the factorial, so the rest can be computed by calculator.

edited Oct 03 '19 at 10:34

answered Oct 03 '19 at 10:24

Toby Mak

16,827

Thanks! Could you also do $26! \sum_{i=0}^{26}\frac{(-1)^i}{i!}$ please? – John B. Oct 03 '19 at 10:30
You can just go to Wolfram Alpha and type it yourself. For the summation, you have to type the limits of the summation out like the end, something like: (sum (-1)^i/i! from i=0 to 26). – Toby Mak Oct 03 '19 at 10:31
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For the summation, it might also be help to note that it very nearly equals $\frac{1}{e}$. – Toby Mak Oct 03 '19 at 10:38

Calculating $\sum_{k=0}^{13} \binom{26}{2k}\frac{(2k)!}{k!2^k}$

1 Answers1