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+1/sqrt2)^200`.If `("^200C_99K)/a=(2^lm)/n` where m and n are odd numbers, then the ordered pair (l, n) is equal to

Options

A. (51, 99)
B. (50, 101)
C. (50, 51)
D. (51, 101)

Answer

To solve the given problem, we will break it down into manageable steps.

### Step 1: Find K
K is defined as the sum of the coefficients of the odd powers of \( x \) in the expansion of \( (1+x)^{99} \).

The sum of the coefficients of the expansion can be found by substituting \( x = 1 \) and \( x = -1 \):
\[ (1+1)^{99} = 2^{99} \quad \text{(sum of all coefficients)} \]
\[ (1-1)^{99} = 0 \quad \text{(sum of coefficients of even powers)} \]

The sum of the coefficients of the odd powers is given by:
\[ \text{Sum of odd coefficients} = \frac{(1+1)^{99} - (1-1)^{99}}{2} = \frac{2^{99} - 0}{2} = 2^{98} \]
Thus, \( K = 2^{98} \).

### Step 2: Find the middle term \( a \) in the expansion of \( \left(2 + \frac{1}{\sqrt{2}}\right)^{200} \)
The middle term in the expansion of \( (a+b)^n \) is given by:
\[ T_{k+1} = \binom{n}{k} a^{n-k} b^k \]
where \( k = \frac{n}{2} \) when \( n \) is even.

Here, \( n = 200 \), so the middle term is:
\[ T_{101} = \binom{200}{100} \left(2\right)^{100} \left(\frac{1}{\sqrt{2}}\right)^{100} \]
Calculating this gives:
\[ T_{101} = \binom{200}{100} \cdot 2^{100} \cdot \left(\frac{1}{\sqrt{2}}\right)^{100} = \binom{200}{100} \cdot 2^{100} \cdot 2^{-50} = \binom{200}{100} \cdot 2^{50} \]
Thus, \( a = \binom{200}{100} \cdot 2^{50} \).

### Step 3: Calculate \( \frac{200C99 \cdot K}{a} \)
Now we need to compute:
\[ \frac{\binom{200}{99} \cdot K}{a} \]
Substituting the values we found:
\[ = \frac{\binom{200}{99} \cdot 2^{98}}{\binom{200}{100} \cdot 2^{50}} \]
This simplifies to:
\[ = \frac{\binom{200}{99}}{\binom{200}{100}} \cdot 2^{48} \]
Using the property of combinations:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
we have:
\[ \frac{\binom{200}{99}}{\binom{200}{100}} = \frac{100}{101} \]
Thus:
\[ \frac{200C99 \cdot K}{a} = \frac{100}{101} \cdot 2^{48} \]

### Step 4: Express in the form \( \frac{2^{lm}}{n} \)
We need to express \( \frac{100}{101} \cdot 2^{48} \) in the form \( \frac{2^{lm}}{n} \):
\[ \frac{100}{101} \cdot 2^{48} = \frac{2^{48} \cdot 100}{101} \]
Here, \( 100 = 2^2 \cdot 25 \), so:
\[ = \frac{2^{50}}{101} \]
This gives us \( l = 50 \) and \( n = 101 \).

### Final Answer
The ordered pair \( (l, n) \) is:
\[ \boxed{(50, 101)} \]