Let be the sum of the coefficients of the odd powers of in the expansion of . Let be the middle term in the expansion of . If , where and are odd numbers, then the ordered pair is equal to [2023]

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Solution

Q.
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 [2023]

1 (50, 101)

2 (51, 101)

3 (51, 99)

4 (50, 51)

Ans.
(1)

$Sum of the coefficients of odd powers of x in the expansion {(1 + x)}^{99} be K .$

$If a be the middle term in expansion of {(2 + \frac{1}{\sqrt{2}})}^{200} .$

$In the expansion of {(1 + x)}^{99} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{99} x^{99}$

$If a = middle term of$ ${(2 + \frac{1}{\sqrt{2}})}^{200}$

$= T_{(\frac{200}{2} + 1)} = {}^{200}C_{100} {(2)}^{100} {(\frac{1}{\sqrt{2}})}^{100} \Rightarrow T_{101} = {}^{200}C_{100} \cdot 2^{50}$

$K = 2^{98}; T_{101} = {}^{200}C_{100} \cdot 2^{50} = a$

$So, \frac{{}^{200}C_{99} \times 2^{98}}{{}^{200}C_{100} \times 2^{50}} = \frac{100}{101} \times 2^{48} \Rightarrow \frac{25}{101} \times 2^{50} = \frac{m}{n} 2^{l}$

$∴$ $m, n are odd, so (l, n) becomes (50, 101)$

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