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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 \left(2+\frac{1}{\sqrt{2}}\right)^{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

Option: 1

(50,51)


Option: 2

(50,101)


Option: 3

(51,99)


Option: 4

(51,101)


Answers (1)

best_answer

\begin{aligned} & \mathrm{K}=\frac{(1+1)^{99}}{2}=2^{98} \\ & \mathrm{a}={ }^{200} \mathrm{C}_{100} 2^{100} \times \frac{1}{(\sqrt{2})^{100}} \\ & \mathrm{a}={ }^{200} \mathrm{C}_{100} 2^{50} \\ & \frac{{ }^{200} \mathrm{C}_{99} \cdot \mathrm{K}}{\mathrm{a}}=\frac{{ }^{200} \mathrm{C}_{99} \cdot 2^{98}}{{ }^{200} \mathrm{C}_{100} \cdot 2^{50}} \\ & \because \frac{{ }^{200} \mathrm{C}_{99}}{{ }^{200} \mathrm{C}_{100}}=\frac{1200}{[99 \mid 101} \times \frac{\lfloor 100\lfloor 100}{\lfloor 200} \end{aligned}

\begin{aligned} & =\frac{100}{101} \\ & \therefore \frac{{ }^{200} \mathrm{C}_{99} \mathrm{~K}}{\mathrm{a}}=\frac{100}{101} \times 2^{48} \\ & =\frac{25 \times 2^{50}}{101} \\ & \ell=50, \mathrm{n}=101 \\ & (\ell, \mathrm{n})=(50,101) \\ & \end{aligned}

 

 

 

Posted by

vishal kumar

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