$$ \text { If } \sum\limits_{k=1}^{10} K^{2}\left(10_{C_{K}}\right)^{2}=22000 L \text {, then } L \text { is equal to }$$ ________.
Answer (integer)
221
Solution
<p>Given,</p>
<p>$\sum\limits_{k = 1}^{10} {{k^2}{{\left( {{}^{10}{C_k}} \right)}^2} = 2200\,L}$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{}^{10}{C_k}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {k\,.\,{{10} \over k}\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow \sum\limits_{k = 1}^{10} {{{\left( {10\,.\,{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow 100\,.\,\sum\limits_{k = 1}^{10} {{{\left( {{}^9{C_{k - 1}}} \right)}^2} = 22000\,L} $$</p>
<p>$$ \Rightarrow 100\left( {{{\left( {{}^9{C_0}} \right)}^2} + {{\left( {{}^9{C_1}} \right)}^2}\, + \,....\, + \,{{\left( {{}^9{C_9}} \right)}^2}} \right) = 22000\,L$$</p>
<p>$\Rightarrow 100\left( {{}^{18}{C_9}} \right) = 22000\,L$</p>
<p>[Note : $${\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2}\, + \,....\, + \,{\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}$$]</p>
<p>$\Rightarrow 100 \times {{18!} \over {9!\,9!}} = 22000\,L$</p>
<p>$\Rightarrow L = 221$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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