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+12200. If 200C99Ka=2lmn,where mand n are odd numbers, then the ordered pair l, n is

k=f(1)−f(−1)2=2992=298a=200C100(2)10012100=200C100250 200C99⋅298 200C100⋅250=2lmn⇒100200−100+1⋅248=2lmn⇒100101⋅248=2lmn⇒25⋅250101=2lmn⇒l=50,n=101

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+12200. If 200C99Ka=2lmn,where mand n are odd numbers, then the ordered pair l, n is

k=f(1)−f(−1)2=2992=298a=200C100(2)10012100=200C100250 200C99⋅298 200C100⋅250=2lmn⇒100200−100+1⋅248=2lmn⇒100101⋅248=2lmn⇒25⋅250101=2lmn⇒l=50,n=101

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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

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$(50, 51)$

$(51, 101)$

$(50, 101)$

$(51, 99)$

answer is B.

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Detailed Solution

$k = \frac{f (1) - f (- 1)}{2} = \frac{2^{99}}{2} = 2^{98}$
$\begin{array}{l} a =^{200} C_{100} (2)^{100} {(\frac{1}{\sqrt{2}})}^{100} =^{200} C_{100} 2^{50} \\ \frac{^{200} C_{99} \cdot 2^{98}}{^{200} C_{100} \cdot 2^{50}} = \frac{2^{l} m}{n} \Rightarrow \frac{100}{200 - 100 + 1} \cdot 2^{48} = \frac{2^{l} m}{n} \\ \Rightarrow \frac{100}{101} \cdot 2^{48} = \frac{2^{l} m}{n} \Rightarrow \frac{25 \cdot 2^{50}}{101} = \frac{2^{l} m}{n} \Rightarrow l = 50, n = 101 \end{array}$