The sum _r=1^10(r^2+1)(r!) is equal to:

Question

The sum $\displaystyle\sum_{r=1}^{10}(r^2+1)	imes(r!)$ is equal to:

Accepted Answer

$$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)}  \times r!$$

$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!} $$

$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]} $$

$$ = \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} } $$

$$ = \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]}  - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!} $$

$$ = \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]$$

$$ - \sum\limits_{r = 1}^{10} {r\,.\,r!} $$

$=$   11.11! - 1.1! - $$\sum\limits_{r = 1}^{10} {r\,.\,r!} $$

$=$ (11.11! - 1) - $$\sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right)} \,.\,r!$$

$=$ (11.11! - 1) - $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)r! - r!} \right]} $$

$=$ (11.11! - 1) - $$\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)! - r!} \right]} $$

$=$ (11.11! - 1) - [(2! - 1!) + (3! - 2!) + . . . .+ (11! - 10!)]

$=$ (11.11! - 1) - (11! - 1)

$=$ 11.11! - 11!

$=$ 11! (11 - 1)

$=$ $$10\,.\,\left( {11!} \right)$$

Solutions · 1