_r=1^20(r^2+1)(r!) is equal to:

Question

$\displaystyle\sum_{r=1}^{20}(r^2+1)(r!)$ is equal to:

Accepted Answer

Given,

$$\sum\limits_{r = 1}^{20} {({r^2} + 1)(r!)} $$

Let, $f(r) = ({r^2} + 1)(r!)$

$= ({r^2})(r!) + r!$

$$ = r(r\,r!) + r!$$

$$ = r[(r + 1 - 1)r!] + r!$$

$$ = r[(r + 1)r! - r!] + r!$$

$$ = r[(r + 1)! - (r!)] + r!$$

$= r(r + 1)! - r(r!) + r!$

$= (r + 2 - 2)(r + 1)! - r(r!) + r!$

$$ = (r + 2)(r + 1)! - 2(r + 1)! - [(r + 1 - 1)(r!)] + r!$$

$$ = (r + 2)! - 2(r + 1)! - (r + 1)! + r! + r!$$

$= (r + 2)! - 3(r + 1)! + 2r!$

$$ = [(r + 2)! - (r + 1)!] - 2[(r + 1)! - r!]$$

$$\therefore$$ $$\sum\limits_{r = 1}^{20} {f(r)} $$

$$ = \sum\limits_{r = 1}^{20} {[(r + 2)! - (r + 1)!] - 2\sum\limits_{r = 1}^{20} {[(r + 1)! - r!]} } $$

$$ = [(22! + 21! + 20!\, + \,.....\, + \,4! + 3!) - (21! + 20! + 19!\, + \,....\, + \,3! + 2!] - 2[(21! + 20!\, + \,.....\, + \,3! + 2!) - (20! + 19!\, + \,.....\,1!)]$$

$$ = [(22!) - (2!)] - 2[(21)! - (1!)]$$

$$ = 22! - 2! - 2\,.\,(21)! + 2\,.\,1!$$

$$ = 22! - 2\,.\,(21)!$$

Solutions · 1