indefinite integral.....

2) U_n = \int_0^1 x^n (2-x)^n \ dx \Rightarrow \frac{U_n}{2^{2n}} = \int_0^1 \left(\frac{x}{2}\right) ^n \left(1-\frac{x}{2} \right)^n \ dx

= 2\int_0^ {\frac{1}{2}} t ^n \left(1-t \right)^n \ dt ; t = \frac{x}{2}

= \int_0^ {\frac{1}{2}} t ^n \left(1-t \right)^n \ dt +\int_0^ {\frac{1}{2}} t ^n \left(1-t \right)^n \ dt

The second integrand by the transformation t \rightarrow 1-t can be converted to

\int_{\frac{1}{2}}^ 1 t ^n \left(1-t \right)^n \ dt

Thus

\frac{U_{2n}}{2^{2n}} = \int_0^{\frac{1}{2}} t ^n \left(1-t \right)^n \ dt + \int_{\frac{1}{2}}^ 1 t ^n \left(1-t \right)^n \ dt = \int_0^1 t ^n \left(1-t \right)^n \ dt = V_n

and we are done

http://targetiit.com/iit-jee-forum/posts/prob-17742.html

see if this helps.