JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 8)
The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to :
(11)!
10 $$ \times $$ (11!)
101 $$ \times $$ (10!)
11 $$ \times $$ (11!)
Explanation
$$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times r!$$
$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!} $$
$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]} $$
$$ = \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} } $$
$$ = \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]} - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!} $$
$$ = \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]$$
$$ - \sum\limits_{r = 1}^{10} {r\,.\,r!} $$
$$=$$ 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)
$$=$$ $$10\,.\,\left( {11!} \right)$$
$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!} $$
$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]} $$
$$ = \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} } $$
$$ = \sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,r\,\,.\,\,r!} \right]} - \sum\limits_{r = 1}^{10} {r\,\,.\,\,r!} $$
$$ = \left[ {\left( {2.2! - 1.1!} \right) + \left( {3.3! - 2.2!} \right) + .... + \left( {11.11! - 10.10!} \right)} \right]$$
$$ - \sum\limits_{r = 1}^{10} {r\,.\,r!} $$
$$=$$ 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)
$$=$$ $$10\,.\,\left( {11!} \right)$$
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