1. Key Concepts: Binomial Theorem and Sum of Coefficients

The problem requires us to find the sum of coefficients of terms with integral powers of $x$ in a binomial expansion. The Binomial Theorem states that for any non-negative integer $n$, $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ A crucial property used here is that for a polynomial $P(t) = a_0 + a_1 t + a_2 t^2 + \dots + a_n t^n$,

• The sum of all coefficients is $P(1) = a_0 + a_1 + \dots + a_n$.
• The sum of coefficients of even powers of $t$ ($a_0 + a_2 + a_4 + \dots$) is $\frac{P(1) + P(-1)}{2}$.
• The sum of coefficients of odd powers of $t$ ($a_1 + a_3 + a_5 + \dots$) is $\frac{P(1) - P(-1)}{2}$.

2. Binomial Expansion and Identifying Integral Powers of x

Let's consider the given expression: ${\left( {1 - 2\sqrt x } \right)^{50}}$ Using the Binomial Theorem with $a=1$, $b=-2\sqrt x$, and $n=50$, the general term (T_k+1) in the expansion is: $T_{k+1} = \binom{50}{k} (1)^{50-k} (-2\sqrt x)^k = \binom{50}{k} (-2)^k (\sqrt x)^k = \binom{50}{k} (-2)^k x^{k/2}$ Here, $k$ ranges from 0 to 50.

We are looking for terms where the power of $x$ ($k/2$) is an integer. For $k/2$ to be an integer, $k$ must be an even number. So, the terms with integral powers of $x$ correspond to $k = 0, 2, 4, \dots, 50$.

Let's list some of these terms and their coefficients:

• For $k=0$: $T_1 = \binom{50}{0} (-2)^0 x^{0/2} = 1 \cdot 1 \cdot x^0 = 1$ (Coefficient = 1)
• For $k=2$: $T_3 = \binom{50}{2} (-2)^2 x^{2/2} = \binom{50}{2} \cdot 4 \cdot x^1$ (Coefficient = $\binom{50}{2} \cdot 4$)
• For $k=4$: $T_5 = \binom{50}{4} (-2)^4 x^{4/2} = \binom{50}{4} \cdot 16 \cdot x^2$ (Coefficient = $\binom{50}{4} \cdot 16$) and so on, up to $k=50$.

The sum we need is the sum of these coefficients: $S = \binom{50}{0}(-2)^0 + \binom{50}{2}(-2)^2 + \binom{50}{4}(-2)^4 + \dots + \binom{50}{50}(-2)^{50}$

3. Applying the Sum of Even-Powered Coefficients Trick

To efficiently sum these specific coefficients, we can use the property for even-powered terms. Let $P(t) = (1-2t)^{50}$. The expansion of $P(t)$ is $P(t) = \sum_{k=0}^{50} \binom{50}{k} (1)^{50-k} (-2t)^k = \sum_{k=0}^{50} \binom{50}{k} (-2)^k t^k$. Notice that the coefficients of $t^k$ in $P(t)$ are $\binom{50}{k}(-2)^k$. These are exactly the coefficients we need for $k=0, 2, 4, \dots, 50$.

Therefore, the sum of coefficients of integral powers of $x$ is equivalent to the sum of coefficients of even powers of $t$ in the expansion of $P(t) = (1-2t)^{50}$.

Using the formula for the sum of coefficients of even powers: $\frac{P(1) + P(-1)}{2}$.

First, calculate $P(1)$: $P(1) = (1 - 2(1))^{50} = (1 - 2)^{50} = (-1)^{50} = 1$

Next, calculate $P(-1)$: $P(-1) = (1 - 2(-1))^{50} = (1 + 2)^{50} = (3)^{50}$

Now, substitute these values into the formula: Sum of coefficients (including $x^0$) $= \frac{P(1) + P(-1)}{2} = \frac{1 + 3^{50}}{2}$

4. Addressing the Nuance: Excluding the Constant Term

The term "integral power of $x$" typically includes $x^0$, as 0 is an integer. Our calculation above ($\frac{1 + 3^{50}}{2}$) includes the coefficient of $x^0$. However, in many competitive exams like JEE, "integral powers of $x$" might implicitly refer to terms that actively involve $x$ (i.e., $x^1, x^2, \dots$), effectively excluding the constant term ($x^0$). Given the options and the provided correct answer, it's highly probable that the constant term ($x^0$) is meant to be excluded from the sum.

The coefficient of $x^0$ (which occurs when $k=0$) is $\binom{50}{0} (-2)^0 = 1 \cdot 1 = 1$.

To get the sum of coefficients of integral powers of $x$ excluding the constant term: Required Sum $= \left( \text{Sum of coefficients including } x^0 \right) - \left( \text{Coefficient of } x^0 \right)$ Required Sum $= \frac{1 + 3^{50}}{2} - 1$ Required Sum $= \frac{1 + 3^{50} - 2}{2}$ Required Sum $= \frac{3^{50} - 1}{2}$

This matches option (A).

5. Tips and Common Mistakes

• Ambiguity of "Integral Power": Be aware that "integral power of $x$" can sometimes be ambiguous. In a general mathematical context, it includes $x^0$. In competitive exams, it might sometimes implicitly exclude $x^0$ if the options suggest so. Always check the options provided.
• Correct Variable Substitution: Ensure you correctly identify what plays the role of '$t$' in the sum of even/odd coefficients formula. Here, the terms were functions of $\sqrt x$, so letting $t = 2\sqrt x$ (or just using the function $P(t)=(1-2t)^{50}$ directly) was key.
• Careful with $(-1)^n$: Remember that $(-1)^{\text{even}} = 1$ and $(-1)^{\text{odd}} = -1$. This was crucial for $(-1)^{50} = 1$.

6. Summary

We expanded the given binomial expression ${\left( {1 - 2\sqrt x } \right)^{50}}$ and identified that integral powers of $x$ correspond to even values of $k$ in the general term $\binom{50}{k} (-2)^k x^{k/2}$. We then recognized that the sum of these coefficients is equivalent to the sum of coefficients of even powers in the polynomial $P(t) = (1-2t)^{50}$. Using the formula $\frac{P(1) + P(-1)}{2}$, we found the sum to be $\frac{1 + 3^{50}}{2}$. Finally, interpreting "integral power of $x$" to exclude the constant term $x^0$, we subtracted its coefficient (which is 1) to arrive at the final answer: $\frac{3^{50} - 1}{2}$.

The final answer is $\boxed{\frac{1}{2}\left( {{3^{50}} - 1} \right)}$.