2n+1C12n+1C1 + 2n+1C22n+1C2 + ... 2n+1Cn−12n+1Cn−1 + 2n+1Cn2n+1Cn = 63 ......... How to solve this equation?

there is no formula? we jus use hit n trial?

Sum is 63 so N should not be above 4. There are just 3 values to be checked. i.e 2,3 and 4. On checking we can see that n=3 satisfies this equation.
Now if you want to know that why I said it won't be above 4, because if you put n=4, you will get 9c1, 9c2, 9c3 and 9c4. 9c4 it self is greater than 63. So N has to be below 4.
Thanks