KVPY #5

${\left( 1 - 2\sqrt{x} \right)}^{50} = \dbinom{50}{0} + \dbinom{50}{1} (-2 \sqrt{x}) + \dbinom{50}{2} (-2 \sqrt{x})^2 + \dbinom{50}{3} (-2 \sqrt{x})^3 + \cdots + \dbinom{50}{49} (-2 \sqrt{x})^{49} + \dbinom{50}{50} (-2 \sqrt{x})^{50}$

We observe that all the even powers of $-2 \sqrt{x}$ will give us a term with an integral power of $x$ . So we need to remove all the odd powers of $-2 \sqrt{x}$ .

Notice that

${\left( 1 + 2\sqrt{x} \right)}^{50} = \dbinom{50}{0} + \dbinom{50}{1} (2 \sqrt{x}) + \dbinom{50}{2} (2 \sqrt{x})^2 + \dbinom{50}{3} (2 \sqrt{x})^3 + \cdots + \dbinom{50}{49} (2 \sqrt{x})^{49} + \dbinom{50}{50} (2 \sqrt{x})^{50}$

If we add the above two binomial expansions, we see that all the terms with non-integral powers of $x$ get cancelled and we get a polynomial in terms of $x$ (remember a polynomial is synonymous with an expansion having only positive integral powers of $x$ ).

Thus

${\left( 1 - 2\sqrt{x} \right)}^{50} + {\left( 1 + 2\sqrt{x} \right)}^{50} = 2 \displaystyle \sum_{n=0}^{25} \dbinom{50}{2n} {\left( \pm 2 \sqrt{x} \right)}^{2n} = 2 \displaystyle \sum_{n=0}^{25} \dbinom{50}{2n} {\left(4x\right)}^n$

So sum of coefficients of all integral powers is the value of $\dfrac{{\left( 1 - 2\sqrt{x} \right)}^{50} + {\left( 1 + 2\sqrt{x} \right)}^{50}}{2}$ at $x=1$ .

Therefore, required sum is

$S = \dfrac{{\left( 1 - 2\sqrt{1} \right)}^{50} + {\left( 1 + 2\sqrt{1} \right)}^{50}}{2} = \dfrac{ (-1)^{50} + 3^{50} }{2} = \boxed{0.5 \left( 3^{50} + 1 \right)}$