The sum of coefficients of integral powers of $$x$$ in the binomial expansion of $$(1 - 2\sqrt{x})^{50}$$ is

We have to add up the coefficients of only those terms in the expansion of $$(1-2\sqrt{x})^{50}$$ that contain an integral (whole-number) power of $$x$$.

First recall the binomial theorem, which states that for any real numbers $$a$$ and $$b$$ and any non-negative integer $$n$$, we have

$$ (a+b)^n \;=\; \sum_{k=0}^{n} \binom{n}{k}\,a^{\,n-k}\,b^{\,k}. $$

To expand $$(1-2\sqrt{x})^{50}$$ we put $$a=1$$ and $$b=-2\sqrt{x}$$, giving the general (the $$k^{\text{th}}$$) term

$$ T_k \;=\; \binom{50}{k}\,(1)^{\,50-k}\,(-2\sqrt{x})^{k} \;=\; \binom{50}{k}\,(-1)^{k}\,2^{k}\,x^{k/2}. $$

The power of $$x$$ that appears in this term is $$k/2$$. For this power to be an integer, $$k/2$$ must itself be an integer, which is possible only when $$k$$ is even. So we set $$k=2m$$, where $$m$$ runs over all integers from $$0$$ to $$25$$:

$$ k = 2m,\qquad m = 0,1,2,\dots ,25. $$

Substituting $$k=2m$$ into the expression for $$T_k$$ gives

$$ T_{2m} \;=\; \binom{50}{2m}\, (-1)^{2m}\,2^{2m}\,x^{m}. $$

Because $$(-1)^{2m}=1$$ for every integer $$m$$, the coefficient of $$x^{m}$$ in this term is simply

$$ \binom{50}{2m}\,2^{2m}. $$

The required sum (call it $$S$$) is the sum of these coefficients for all admissible values of $$m$$:

$$ S \;=\; \sum_{m=0}^{25} \binom{50}{2m}\,2^{2m}. $$

To evaluate this quickly, we use a standard identity for separating even and odd terms of a binomial expansion.

First write the expansion of $$(1+2)^{50}$$ and $$(1-2)^{50}$$ using the binomial theorem:

$$ (1+2)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,1^{\,50-k}\,2^{k} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,2^{k}, $$

$$ (1-2)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,1^{\,50-k}\,(-2)^{k} \;=\; \sum_{k=0}^{50} (-1)^{k}\,\binom{50}{k}\,2^{k}. $$

Now add these two expansions term by term:

$$ (1+2)^{50} + (1-2)^{50} \;=\; \sum_{k=0}^{50} \binom{50}{k}\,2^{k} + \sum_{k=0}^{50} (-1)^{k}\,\binom{50}{k}\,2^{k}. $$

For every odd $$k$$, the two corresponding terms are equal in magnitude but opposite in sign and therefore cancel. For every even $$k$$, the signs are the same and the terms double. Hence the sum simplifies to

$$ (1+2)^{50} + (1-2)^{50} \;=\; 2 \sum_{k\;\text{even}} \binom{50}{k}\,2^{k} \;=\; 2 \sum_{m=0}^{25} \binom{50}{2m}\,2^{2m}. $$

Comparing with our definition of $$S$$, we see that

$$ 2S \;=\; (1+2)^{50} + (1-2)^{50}. $$

But $$1+2 = 3$$ and $$1-2 = -1$$, so

$$ (1+2)^{50} = 3^{50},\qquad (1-2)^{50} = (-1)^{50} = 1. $$

Therefore

$$ 2S \;=\; 3^{50} + 1, $$

and dividing by $$2$$ gives

$$ S \;=\; \frac{3^{50} + 1}{2}. $$

This expression is exactly option B.

Hence, the correct answer is Option B.