$\sum\limits_{r=1}^{20}\left(r^{2}+1\right)(r !)$ is equal to
Solution
<p>Given,</p>
<p>$\sum\limits_{r = 1}^{20} {({r^2} + 1)(r!)}$</p>
<p>Let, $f(r) = ({r^2} + 1)(r!)$</p>
<p>$= ({r^2})(r!) + r!$</p>
<p>$= r(r\,r!) + r!$</p>
<p>$= r[(r + 1 - 1)r!] + r!$</p>
<p>$= r[(r + 1)r! - r!] + r!$</p>
<p>$= r[(r + 1)! - (r!)] + r!$</p>
<p>$= r(r + 1)! - r(r!) + r!$</p>
<p>$= (r + 2 - 2)(r + 1)! - r(r!) + r!$</p>
<p>$= (r + 2)(r + 1)! - 2(r + 1)! - [(r + 1 - 1)(r!)] + r!$</p>
<p>$= (r + 2)! - 2(r + 1)! - (r + 1)! + r! + r!$</p>
<p>$= (r + 2)! - 3(r + 1)! + 2r!$</p>
<p>$= [(r + 2)! - (r + 1)!] - 2[(r + 1)! - r!]$</p>
<p>$\therefore$ $\sum\limits_{r = 1}^{20} {f(r)}$</p>
<p>$$ = \sum\limits_{r = 1}^{20} {[(r + 2)! - (r + 1)!] - 2\sum\limits_{r = 1}^{20} {[(r + 1)! - r!]} } $$</p>
<p>$$ = [(22! + 21! + 20!\, + \,.....\, + \,4! + 3!) - (21! + 20! + 19!\, + \,....\, + \,3! + 2!] - 2[(21! + 20!\, + \,.....\, + \,3! + 2!) - (20! + 19!\, + \,.....\,1!)]$$</p>
<p>$= [(22!) - (2!)] - 2[(21)! - (1!)]$</p>
<p>$= 22! - 2! - 2\,.\,(21)! + 2\,.\,1!$</p>
<p>$= 22! - 2\,.\,(21)!$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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