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. 
A: Following on user94529's answer, since the sum $\sum_k \binom{2n}{2k+1} = 2^{2n-1}$ is a power of two, we will be done if we look only at the greatest power of two dividng them all. By experimenting a bit, it seems that $\binom{2n}{1} = 2n$ has the smallest power of 2 of all. 
This is indeed the case, looking at $\binom{2n}{2k+1} = \frac{2n}{2k+1} \times \binom{2n-1}{2k}$ since $2k+1$ is odd and $\binom{2n-1}{2k}$ is an integer, we get that the greatest power of 2 dividing $\binom{2n}{2k+1}$ is greater or equal to that of $2n$.
Hence, if $n=2^j \times m$ where $m$ is odd, then the answer is
$$ \gcd\left( \binom{2n}{1},\ldots,\binom{2n}{2n-1} \right) = 2^{j+1} $$
A: As user94529 showed,
the gcd is a power of 2.
Looking just at the first term,
the gcd must divide $2n$.
In particular,
if $n$ is odd,
the gcd is 2 or 1.
I now use the answer to this question:
Odd Binomial Coefficients?
The answer by Arturo Magidin
says
"This follows easily from Kummer's Theorem, that the highest power of a prime $p$ that divides $\binom{n}{m}$ is equal to the number of "carries" when adding $n-m$ and $m$ in base $p$. In particular, $\binom{n}{m}$ is odd if and only if there are no carries when adding in base $2$."
Both $2k-1$ and $2n-(2k-1)$ are odd,
so there is a carry when they are added in base 2.
Therefore,
all of
$\binom{2n}{2k-1}$ are even.
Therefore, if $n$ is odd,
the gcd is 2.
Perhaps by looking at the
power of 2 than divides $n$
we could get a result for all $n$.
