Hard MCQ +4 / -1 PYQ · JEE Mains 2022

If

$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $$,

where $\alpha$ $\in$ R, then the value of 16$\alpha$ is equal to

  1. A 1411 Correct answer
  2. B 1320
  3. C 1615
  4. D 1855

Solution

<p>Given,</p> <p>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } $$</p> <p>Now,</p> <p>$$\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right)} $$</p> <p>$$ = \left( {{}^{31}{C_1}\,.\,{}^{31}{C_0} + {}^{31}{C_2}\,.\,{}^{31}{C_1} + {}^{31}{C_3}\,.\,{}^{31}{C_2} + \,\,......\,\, + \,\,{}^{31}{C_{31}}\,.\,{}^{31}{C_{30}}} \right)$$</p> <p>$$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{31 - 1}} + {}^{31}{C_1}\,.\,{}^{31}{C_{31 - 2}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_{31 - 31}}} \right)$$</p> <p>[using ${}^n{C_r} = {}^n{C_{n - r}}$]</p> <p>$$ = \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{30}} + {}^{31}{C_1}\,.\,{}^{31}{C_{29}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_0}} \right)$$</p> <p>$= {}^{62}{C_{30}}$</p> <p>Now, $\sum\limits_{k = 1}^{30} {{}^{30}{C_k}\,.\,{}^{30}{C_{k - 1}}}$</p> <p>$$ = \left( {{}^{30}{C_1}\,.\,{}^{30}{C_0} + {}^{30}{C_2}\,.\,{}^{30}{C_1} + \,\,.....\,\, + \,\,{}^{30}{C_{30}}\,.\,{}^{30}{C_{29}}} \right)$$</p> <p>$$ = \left( {{}^{30}{C_0}\,.\,{}^{30}{C_{29}} + {}^{30}{C_1}\,.\,{}^{30}{C_{28}} + \,\,....\,\, + \,\,{}^{30}{C_{29}}\,.\,{}^{30}{C_0}} \right)$$</p> <p>$= {}^{60}{C_{29}}$</p> <p>$\therefore$ ${}^{60}{C_{30}} - {}^{60}{C_{29}} = {{\alpha (60!)} \over {30!\,31!}}$</p> <p>$$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {29!\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$$</p> <p>$$ \Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {{{30!} \over {30}}\,.\,31!}} = {{\alpha (60!)} \over {30!\,31!}}$$</p> <p>$$ \Rightarrow {{60!} \over {30!\,31!}}\left( {{{62\,.\,61} \over {32}} - 30} \right) = {{\alpha (60!)} \over {30!\,31!}}$$</p> <p>$\Rightarrow \alpha = {{62\,.\,61} \over {32}} - 30$</p> <p>$\Rightarrow 16\alpha = {{62\,.\,61 - 30 \times 32} \over 2}$</p> <p>$\Rightarrow 16\alpha = {{2822} \over 2} = 1411$</p>

About this question

Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion

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