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QJEEM16Ma0d1a: JEE Main 2016 Mathematics - Class 11 JEEM16Ma0d1a Solution: The sum is equal to :

The sum ∑r=110(r2+1)×(r!)\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} is equal to :

Multiple Choice Options

  • A (11)!
  • B 10 ×\times (11!)
  • C 101 ×\times (10!)
  • D 11 ×\times (11!)

Step-by-Step Verified Solution

∑r=110(r2+1)×r!\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times r!

=∑r=110[(r+1)2−2r] . r!= \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!}

=∑r=110[(r+1)2 . r!  −  2r . r!]= \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]}

=∑r=110(r+1)(r+1)!  −  2∑r=110r . r!= \sum\limits_{r = 1}^{10} {\left( {r + 1} \right)\left( {r + 1} \right)!\,\, - \,\,2\sum\limits_{r = 1}^{10} {r\,.\,r!} }

=∑r=110[(r+1)(r+1)!  − r  .  r!]−∑r=110r  .  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!}

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

       −∑r=110r . r!- \sum\limits_{r = 1}^{10} {r\,.\,r!}

==    11.11! −- 1.1! −- ∑r=110r . r!\sum\limits_{r = 1}^{10} {r\,.\,r!}

== (11.11! −- 1) −- ∑r=110(r+1−1) . r!\sum\limits_{r = 1}^{10} {\left( {r + 1 - 1} \right)} \,.\,r!

== (11.11! −- 1) −- ∑r=110[(r+1)r!−r!]\sum\limits_{r = 1}^{10} {\left[ {\left( {r + 1} \right)r! - r!} \right]}

== (11.11! −- 1) −- ∑r=110[(r+1)!−r!]\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 . (11!)10\,.\,\left( {11!} \right)

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