Medium MCQ +4 / -1 PYQ · JEE Mains 2021

$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}}$ is equal to :

  1. A ${}^{40}{C_{21}}$
  2. B ${}^{40}{C_{19}}$
  3. C ${}^{40}{C_{20}}$ Correct answer
  4. D ${}^{41}{C_{20}}$

Solution

$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}}$ <br><br>= $${\left( {{}^{20}{C_0}} \right)^2} + {\left( {{}^{20}{C_1}} \right)^2} + {\left( {{}^{20}{C_2}} \right)^2} + .... + {\left( {{}^{20}{C_{20}}} \right)^2}$$ <br><br>= <sup>40</sup>C<sub>20</sub> <br><br><b>Using the formula :</b> <br><br>$${\left( {{}^n{C_0}} \right)^2} + {\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2} + .... + {\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}$$

About this question

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

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