Binomial sum

juantheron · Bericht door **juantheron** » 09 mar 2013, 09:43

value of $\displaystyle \frac{\sum_{k=0}^{24}\binom{100}{k}.\binom{100}{4k+2}}{\sum_{k=1}^{25}\binom{200}{8r-6}} =$

where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$

Bericht door **arie** » 10 mar 2013, 17:07

I suppose your r in the binomial in the denominator should be k, so:

$\frac{\sum_{k=0}^{24}\binom{100}{k}.\binom{100}{4k+2}}{\sum_{k=1}^{25}\binom{200}{8k-6}}$

$= \frac{\sum_{k=0}^{24}\binom{100}{k}.\binom{100}{4k+2}}{\sum_{k=0}^{24}\binom{200}{4k+2}}$

but now it gets ugly:

$=\frac{4323748433337035297851365038464157409197847150}{200867255532373784442745261542328412665218316872474928611328}$

and with numerator and denominator decomposed in primes:

$=\frac{2 \cdot 3 \cdot 5 \cdot 11 \cdot 13 \cdot 307 \cdot 611336993 \cdot 1074024205384131858300981209417}{2^{98} \cdot 7 \cdot 23 \cdot 73 \cdot 89 \cdot 199 \cdot 153649 \cdot 599479 \cdot 33057806959}$

Are you sure that the problem statement is correct?
What is the context of this problem?
Do you have some more background information?

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Binomial sum

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