Let K be the sum of the coefficients of the odd powers of x in the expansion of (1+x)⁹⁹. Let a be th

Q: Wait, why is the sum of odd coefficients $2^{n-1}$? I thought it was $2^n$.

That is a classic confusion! The total sum of all coefficients in $(1+x)^n$ is $2^n$ (by setting $x=1$). However, the sum of odd coefficients is exactly half of that total sum because the odd and even coefficients balance each other out in the expansion of $(1+x)^n - (1-x)^n$. Thus, it is $2^{n-1}$.

Q: Why is the middle term $T_{101}$ and not $T_{100}$?

Think about the number of terms. In the expansion of $(x+y)^n$, there are $n+1$ terms. If $n=200$, there are $201$ terms. The middle term is the one that has an equal number of terms before and after it. That is the $101^{st}$ term. So, $r+1 = 101$, which means $r=100$.

Q: How do I know when to stop simplifying the fraction $\frac{2^l m}{n}$?

The condition is that $m$ and $n$ must be odd. This means you must extract every single factor of $2$ from the numerator and denominator. Once you have a power of $2$ multiplied by an odd number in the numerator, and an odd number in the denominator, you have reached the simplest form required by the problem.

Question

Let $K$ be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+x)^{99}$. Let $a$ be the middle term in the expansion of $(2 + \frac{1}{\sqrt{2}})^{200}$. If $\frac{{}^{200}C_{99} K}{a} = \frac{2^l m}{n}$, where $m$ and $n$ are odd numbers, then the ordered pair $(l, n)$ is equal to :

Options:

(A) $(50, 51)$

(B) $(51, 99)$

(C) $(50, 101)$

(D) $(51, 101)$

Sigma Percentile · Accepted Answer

### The Symphony of Binomials: A Journey into Coefficients

Welcome, fellow traveler of the JEE path. Today, we are not just solving a problem; we are uncovering the hidden symmetry within binomial expansions.

This problem is a beautiful test of your ability to manipulate expressions with grace and precision. Let us break it down, step by step, and find the elegance hidden within the algebra.

### Phase 1: The Mystery of $K$

We begin with $K$, the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+x)^{99}$.

Imagine the expansion: 
$$(1+x)^{99} = {}^{99}C_0 + {}^{99}C_1 x + {}^{99}C_2 x^2 + \dots + {}^{99}C_{99} x^{99}$$

We want the sum of coefficients where the power of $x$ is odd. There is a standard, powerful trick here. If we let $f(x) = (1+x)^{99}$, then the sum of all coefficients is $f(1) = 2^{99}$.

The sum of coefficients with alternating signs is $f(-1) = (1-1)^{99} = 0$. By subtracting these two, we isolate the odd terms: 
$$f(1) - f(-1) = 2 	imes (	ext{sum of odd coefficients})$$

Since $f(-1) = 0$, the sum of odd coefficients is simply:
$$\frac{2^{99}}{2} = 2^{98}$$

Thus, **$K = 2^{98}$**. It is a clean, satisfying result, isn't it?

### Phase 2: The Middle Term $a$

Next, we turn our attention to the expansion of $(2 + \frac{1}{\sqrt{2}})^{200}$. We seek the middle term $a$.

Since the exponent $n=200$ is even, there is a unique middle term. This is the $101^{st}$ term, $T_{101}$.

Using the general term formula $T_{r+1} = {}^{n}C_{r} x^{n-r} y^{r}$, we set $r=100$:
$$a = {}^{200}C_{100} (2)^{200-100} \left(\frac{1}{\sqrt{2}}ight)^{100}$$

Now, let us simplify this with care. The term $(\frac{1}{\sqrt{2}})^{100}$ is $(2^{-1/2})^{100} = 2^{-50}$.

So, $a = {}^{200}C_{100} \cdot 2^{100} \cdot 2^{-50}$. Combining the powers of $2$, we get:
$$a = {}^{200}C_{100} \cdot 2^{50}$$

This is the heart of our expression.

### Phase 3: The Ratio and the Final Extraction

Now, we assemble our target expression: $\frac{{}^{200}C_{99} K}{a}$. Substituting our values, we have:
$$\frac{{}^{200}C_{99} \cdot 2^{98}}{{}^{200}C_{100} \cdot 2^{50}}$$

We can separate this into two distinct parts: the ratio of the binomial coefficients and the ratio of the powers of $2$. The powers of $2$ simplify to $2^{98-50} = 2^{48}$.

For the binomial coefficients, we use the beautiful identity:
$$\frac{{}^{n}C_{r-1}}{{}^{n}C_{r}} = \frac{r}{n-r+1}$$

With $n=200$ and $r=100$, this becomes:
$$\frac{100}{200-100+1} = \frac{100}{101}$$

Putting it all together, our expression is $\frac{100}{101} \cdot 2^{48}$.

The problem demands the form $\frac{2^l m}{n}$ where $m$ and $n$ are odd. Our current numerator has $100$, which is $25 	imes 4$, or $25 	imes 2^2$.

So, we write:
$$\frac{25 \cdot 2^2 \cdot 2^{48}}{101} = \frac{25 \cdot 2^{50}}{101}$$

Comparing this to $\frac{2^l m}{n}$, we immediately see that $l=50$, $m=25$, and $n=101$. Both $25$ and $101$ are odd, satisfying our condition perfectly.

The ordered pair $(l, n)$ is **$(50, 101)$**. You have navigated the complexity and arrived at the truth. Keep this clarity with you; it is the key to mastering JEE Advanced.

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Visualized Solution

Finding $K$ : Sum of Odd Coefficients

Finding $a$ : The Middle Term

Simplifying the Middle Term $a$

Setting up the Target Ratio

Simplifying the Powers of $2$

Simplifying Binomial Coefficients

Combining and Extracting Odd Factors

Final Comparison and Result

The Sigma Insight: Properties of Binomial Coefficients

The Symphony of Binomials

Phase 1

Phase 2

Phase 3

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Visualized Solution

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K: Sum of Odd Coefficients

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a: The Middle Term

Simplifying the Middle Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a