Q:

How can we find the gcd for elements (binomial coefficient)?

$\gcd\left(\binom{2n}1,\binom{2n}3,\binom{2n}5,\ldots,\binom{2n}{2n-1}\right)$
i want to know what is specialty of such a series.I am not able to generalize the problem solution.Is there any rule for such gcd computation.

A:

Use -
$$(1+x)^{2n}=\binom{2n}0x^0+\binom{2n}1x^1+\binom{2n}2x^2+\binom{2n}3x^3+\binom{2n}4x^4+...+\binom{2n}{2n}x^{2n}$$
First substituting $ x=-1$ 
$$0=\binom{2n}0-\binom{2n}1+\binom{2n}2...-\binom{2n}{2n-1}+\binom{2n}{2n}$$$$\implies \binom{2n}0+\binom{2n}2+\binom{2n}4+...+\binom{2n}{2n}=\binom{2n}1+\binom{2n}3+\binom{2n}5+...+\binom{2n}{2n-1} $$
But substituting  $x=1$,
we have$$\binom{2n}0+\binom{2n}1+\binom{2n}2+\binom{2n}3+\binom{2n}4+...+\binom{2n}{2n}=2^{2n}$$
$$\implies\binom{2n}1+\binom{2n}3+\binom{2n}5+...+\binom{2n}{2n-1}=2^{2n-1}$$
Hence the gcd is a power of 2..
Now it is sufficient only study the powers of 2 present in the different terms.
This will require some analysis and one way to begin would be assuming$n=\sum2^{k} i.e.$ writing the number in binary form. 

