The sum of coefficients of integral power of x in the binomial expansion (1 - 2√ x)⁵⁰ is (a) (1/2)(3

Q: Wait, why do we use $t = \sqrt{x}$? Can't we just look at the expansion directly?

You absolutely can! However, using a dummy variable like $t = \sqrt{x}$ transforms the expression into a standard polynomial $g(t) = (1 - 2t)^{50}$. This allows us to use powerful algebraic properties for polynomials, like the sum of even-indexed coefficients, which makes the calculation much cleaner and less prone to errors.

Q: How do we know the formula $\frac{P(1) + P(-1)}{2}$ works for the sum of even coefficients?

Think about the expansion $P(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 + \dots$. If you calculate $P(1)$, you get $a_0 + a_1 + a_2 + a_3 + \dots$. If you calculate $P(-1)$, you get $a_0 - a_1 + a_2 - a_3 + \dots$. When you add these two, the odd-indexed terms (like $a_1, a_3, \dots$) cancel out, leaving you with $2(a_0 + a_2 + a_4 + \dots)$. Dividing by two gives you exactly the sum of the even-indexed coefficients!

Q: What if the question asked for the sum of coefficients of odd powers of $x$?

That is a fantastic question! If you wanted the sum of odd coefficients, you would simply subtract the two values instead of adding them: $\frac{P(1) - P(-1)}{2}$. The logic remains the same; it is all about how the signs behave when you plug in $1$ and $-1$.

Question

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

Options:

(A) $\frac{1}{2}(3^{50} - 1)$

(B) $\frac{1}{2}(2^{50} + 1)$

(C) $\frac{1}{2}(3^{50} + 1)$

(D) $\frac{1}{2}(3^{50})$

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Symmetry

Welcome, fellow traveler on the path of JEE mastery. Today, we are going to unravel a problem that, at first glance, might seem like a tedious exercise in expansion, but is actually a beautiful demonstration of algebraic symmetry.

We are looking at the binomial expansion of $(1 - 2\sqrt{x})^{50}$ and seeking the sum of the coefficients of the integral powers of $x$. Let us dive in.

### Deconstructing the Expression

Imagine you are standing before the expression $(1 - 2\sqrt{x})^{50}$. If we were to expand this using the Binomial Theorem, we would have a long string of terms.

The general term in this expansion is given by:

$$T_{k+1} = \binom{50}{k} (1)^{50-k} (-2\sqrt{x})^k = \binom{50}{k} (-2)^k x^{k/2}$$

The question asks for the sum of coefficients of **integral** powers of $x$. For the power of $x$ to be an integer, the exponent $k/2$ must be an integer, which implies that $k$ must be an even number.

This is the geometric reality of our problem: we only care about the terms where $k$ is even.

### The Elegant Substitution

To make this manageable, let us introduce a dummy variable. Let $t = \sqrt{x}$.

Our expression now becomes a polynomial $g(t) = (1 - 2t)^{50}$. The problem has now shifted: we need the sum of the coefficients of the even powers of $t$ in this new polynomial $g(t)$.

This is a classic JEE technique—transforming a complex-looking expression into a standard polynomial form where we can apply known theorems.

### The Magic of Even Coefficients

For any polynomial $P(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 + \dots + a_n t^n$, there is a beautiful property to extract the sum of coefficients of even powers.

If we evaluate the polynomial at $t = 1$, we get $P(1) = a_0 + a_1 + a_2 + a_3 + \dots$. If we evaluate it at $t = -1$, we get $P(-1) = a_0 - a_1 + a_2 - a_3 + \dots$.

Notice what happens when we add these two:

$$P(1) + P(-1) = 2(a_0 + a_2 + a_4 + \dots)$$

Thus, the sum of the even coefficients is simply:

$$	ext{Sum} = \frac{P(1) + P(-1)}{2}$$

### The Final Calculation

Now, let us apply this to our polynomial $g(t) = (1 - 2t)^{50}$. First, we calculate $g(1)$:

$$g(1) = (1 - 2(1))^{50} = (1 - 2)^{50} = (-1)^{50} = 1$$

Next, we calculate $g(-1)$:

$$g(-1) = (1 - 2(-1))^{50} = (1 + 2)^{50} = 3^{50}$$

Finally, we plug these into our formula for the sum of even coefficients:

$$	ext{Sum} = \frac{g(1) + g(-1)}{2} = \frac{1 + 3^{50}}{2}$$

And there you have it! The sum of the coefficients of the integral powers of $x$ is **$\frac{1}{2}(3^{50} + 1)$**.

It is a result that emerges not from brute force, but from understanding the underlying structure of the binomial expansion. Keep this perspective, and you will find that even the most daunting problems have an elegant, hidden path to the solution.

Animated Solution for Mathematics - Binomial Theorem: The sum of coefficients of integral power of $x$ in the binomial expansion $(1 - 2 x)^{50}$ is

Select Answer:

Visualized Solution

Analyze the Expression $(1 - 2 x)^{50}$

Expand the Binomial

Identify Integral Powers

Introduce Dummy Variable $t$

Sum of Even Coefficients Formula

Evaluate $g (1)$

Evaluate $g (- 1)$

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Binomial Symmetry

Deconstructing the Expression

The Elegant Substitution

The Magic of Even Coefficients

The Final Calculation

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of coefficients of integral power of x in the binomial expansion (1−2x​)50 is