Another one of those " try it"

2m?

No , sir , the answer is somewhat very unusual .

m-12 ?

Seems I only have to give the answer ---- :(

2mC1 + 2mC3 + 2mC5 + .............. + 2mC2m-1 = 2^2m-1

As the divisor of each term in l.h.s will be a divisor of r.h.s , so that means that the divisor will be of the form 2^k.

Thus , let the divisor d = 2^k , 0 < k ≤ 2m - 1

Let m = 2ⁿ . r , where r is an odd integer ( any number can be expressed in this form )

So 2m = 2ⁿ⁺¹ . r

So 2mC1 = 2m = 2ⁿ⁺¹ . r

Since d is of the form 2^k and the smallest term 2mC1 = 2ⁿ⁺¹ . r

So d ≤ 2ⁿ⁺¹

Let's say that 2ⁿ⁺¹ divides all the terms .

So for odd positive integer p ,

2mCp

= 2mp . ^2m-1C_p-1

=2ⁿ⁺¹. rp . ^2m-1C_p-1

Now 2mCp is an integer , and p , being odd , doesn't divide 2ⁿ⁺¹.

So we get 2mCp = 2ⁿ⁺¹ . ( an integer )

So 2ⁿ⁺¹ divides 2mCp for all odd values of p starting from 1 to 2m - 1 .

Hence 2ⁿ⁺¹ is the g.c.d of 2mC1 , 2mC3 , ....... when m = 2ⁿ . an odd integer