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

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

Answer 1

The Correct Option is B

In the expansion of \[ (1 + x)^{99} = C_0 + C_1 x + C_2 x^2 + \dots + C_{99} x^{99} \] we define \( K \) as: \[ K = C_1 + C_3 + \dots + C_{99} = 2^{98} \] To find the middle term in the expansion of \[ \left( 2 + \frac{1}{\sqrt{2}} \right)^{200} \] we consider the term: \[ T_{\frac{200}{2} + 1} = C_{100}^{200} (2)^{100} \left( \frac{1}{\sqrt{2}} \right)^{100} \] \[ = C_{100}^{200} \cdot 2^{50} \] Thus, we get: \[ \frac{200}{200} \cdot \frac{C_{99} \times 2^{98}}{C_{100}^{200} \times 2^{50}} = \frac{100}{101} \times 2^{48} \] So, \[ \frac{25}{101} \times 2^{50} = \frac{m}{n} 2^{\ell} \] Since \( m \) and \( n \) are odd, we conclude that: \[ (\ell, n) = (50, 101) \quad \text{Ans.} \] - The problem involves the binomial expansion of \((1 + x)^{99}\) and the summation of alternating binomial coefficients.
- The middle term in the expansion of \(\left( 2 + \frac{1}{\sqrt{2}} \right)^{200}\) is found using the binomial theorem.
- Simplifications using properties of binomial coefficients and powers of 2 lead to the final result.
- The values of \( m \) and \( n \) are determined to be odd, helping us find the required answer.