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
Solution
<p>$K = {2^{98}}$</p>
<p>$a = {}^{200}{C_{100}}\,{2^{50}}$</p>
<p>$\therefore$ $${{{}^{200}{C_{99}}\,.\,{2^{98}}} \over {{}^{200}{C_{100}}\,.\,{2^{50}}}} = {{{2^l}m} \over n}$$</p>
<p>$\Rightarrow {{100} \over {101}}\,.\,{2^{48}} = {{{2^l}m} \over n}$</p>
<p>$\Rightarrow {{25} \over {101}}\,.\,{2^{50}} = {{{2^l}m} \over n}$</p>
<p>$\therefore$ $l = 50,m = 25,n = 101$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Applications of Binomial Theorem
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