Answer:
Option A
Explanation:
Let $T_{r+1}$ be the general term in the expansion of $(1-2\sqrt{x})^{50}$
$\therefore$ $T_{r+1}$ = $^{50}C_{r}(1)^{50-r}(-2x^{1/2})^{r}$
= $^{50}C_{r}2^{r}x^{1/2}(-1)^{r}$
For the integral power of x, r should be even integer.
$\therefore$ Sum of coefficients= $\sum_{r=0}^{25}$ $^{50}C_{2r}(2)^{2r}$
= $\frac{1}{2}[(1+2)^{50}+(1-2)^{50}]$
= $\frac{1}{2}[3^{50}+1]$
Alter:
We have ,
$(1-2\sqrt{x})^{50}=C_{0}-C_{1}2\sqrt{x}+C_{2}(2\sqrt{x})^{2}+...+C_{50}(2\sqrt{x})^{50}......(i)$
$(1+2\sqrt{x})^{50}=C_{0}+C_{1}2\sqrt{x}+C_{2}(2\sqrt{x})^{2}+...+C_{50}(2\sqrt{x})^{50}......(ii)$
On adding Eqs.(i) and (ii) , we get
$(1-2\sqrt{x})^{50}+(1+2\sqrt{x})^{50}=2[C_{0}+C_{2}(2\sqrt{x})^{2}+...+C_{50}(2\sqrt{x})^{50}]$
$\Rightarrow \frac{(1-2\sqrt{x)}^{50}+(1+2\sqrt{x})^{50}}{2}=C_{0}$
$+C_{2}(2\sqrt{x})^{2}+......+C_{50}(2\sqrt{x})^{50}$
On putting x=1, we get
$\frac{(1-2\sqrt{1})^{50}+(1+2\sqrt{1})^{50}}{2}=C_{0}+C_{2}(2)^{2}+...+C_{50}(2)^{50}$
$\Rightarrow \frac{(-1)^{50}+(3)^{50}}{2}=C_{0}+C_{2}(2)^{2}+......+C_{50}(2)^{50}$
$ \Rightarrow \frac{1+3^{50}}{2}=C_{0}+C_{2}(2)^{2}+....+C_{50}(2)^{50}$
