If $$1 + (2 + {}^{49}{C_1} + {}^{49}{C_2} + \,\,...\,\, + \,\,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4} + \,\,...\,\, + \,\,{}^{50}{C_{50}})$$ is equal to $2^{\mathrm{n}} \cdot \mathrm{m}$, where $\mathrm{m}$ is odd, then $\mathrm{n}+\mathrm{m}$ is equal to __________.
Answer (integer)
99
Solution
<p>$$l = 1 + (1 + {}^{49}{C_0} + {}^{49}{C_1}\, + \,....\, + \,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}})$$</p>
<p>As ${}^{49}{C_0} + {}^{49}{C_1}\, + \,.....\, + \,{}^{49}{C_{49}} = {2^{49}}$</p>
<p>and ${}^{50}{C_0} + {}^{50}{C_2}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}}$</p>
<p>$$ \Rightarrow {}^{50}{C_2} + {}^{50}{C_4}\, + \,....\, + \,{}^{50}{C_{50}} = {2^{49}} - 1$$</p>
<p>$\therefore$ $l = 1 + ({2^{49}} + 1)({2^{49}} - 1)$</p>
<p>$= {2^{98}}$</p>
<p>$\therefore$ $m = 1$ and $n = 98$</p>
<p>$m + n = 99$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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