JEE MAIN - Mathematics (2023 - 29th January Evening Shift - No. 12)
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 $${\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}}$$. If $${{{}^{200}{C_{99}}K} \over a} = {{{2^l}m} \over n}$$, where m and n are odd numbers, then the ordered pair $$(l,\mathrm{n})$$ is equal to
(50, 101)
(50, 51)
(51, 101)
(51, 99)
Explanation
$$K = {2^{98}}$$
$$a = {}^{200}{C_{100}}\,{2^{50}}$$
$$\therefore$$ $${{{}^{200}{C_{99}}\,.\,{2^{98}}} \over {{}^{200}{C_{100}}\,.\,{2^{50}}}} = {{{2^l}m} \over n}$$
$$ \Rightarrow {{100} \over {101}}\,.\,{2^{48}} = {{{2^l}m} \over n}$$
$$ \Rightarrow {{25} \over {101}}\,.\,{2^{50}} = {{{2^l}m} \over n}$$
$$\therefore$$ $$l = 50,m = 25,n = 101$$
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