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Permutations and Combinations question

2016 · 10 Apr · Shift 1 · Q28
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Permutations and Combinations question

2016 · 10 Apr · Shift 1 · Q28

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The sum ∑r=110(r2+1)×(r!)\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)}r=1∑10​(r2+1)×(r!) is equal to :
  1. A
    (11)!
  2. B
    10 ×\times× (11!)
  3. C
    101 ×\times× (10!)
  4. D
    11 ×\times× (11!)
View written solutionFree

Correct answer: B

  1. We need to evaluate S=∑r=110(r2+1) r!S=\sum_{r=1}^{10}(r^2+1)\,r!S=∑r=110​(r2+1)r!

  2. Simplify the term: r2+1=r(r−1)+r+1r^2+1=r(r-1)+r+1r2+1=r(r−1)+r+1 So, (r2+1)r!=r(r−1)r!+(r+1)r!(r^2+1)r!=r(r-1)r!+(r+1)r!(r2+1)r!=r(r−1)r!+(r+1)r!

Now use factorial identities:

  • r(r−1)r!=r(r−1)⋅r!=(r+1−1)(r)⋅r!r(r-1)r!=r(r-1)\cdot r!=(r+1-1)(r)\cdot r!r(r−1)r!=r(r−1)⋅r!=(r+1−1)(r)⋅r! A cleaner way is to rewrite directly in terms of factorials: r⋅r!=(r+1)!−r!r\cdot r!=(r+1)!-r!r⋅r!=(r+1)!−r! Hence, r(r−1)r!=(r−1)((r+1)!−r!)r(r-1)r!=(r-1)\big((r+1)!-r!\big)r(r−1)r!=(r−1)((r+1)!−r!) But an even better observation is: r2+1=r(r+1)−r+1r^2+1=r(r+1)-r+1r2+1=r(r+1)−r+1 Thus, (r2+1)r!=r(r+1)r!−r r!+r!(r^2+1)r!=r(r+1)r!-r\,r!+r!(r2+1)r!=r(r+1)r!−rr!+r! Now, r(r+1)r!=(r+1)(r r!)=(r+1)((r+1)!−r!)r(r+1)r!=(r+1)(r\,r!)=(r+1)\big((r+1)!-r!\big)r(r+1)r!=(r+1)(rr!)=(r+1)((r+1)!−r!) This is still not the neatest route.

  1. Use the standard telescoping trick: We try to express (r2+1)r!=(ar+b)(r+1)!+(cr+d)r!(r^2+1)r!=(ar+b)(r+1)!+(cr+d)r!(r2+1)r!=(ar+b)(r+1)!+(cr+d)r! But the simplest identity is: (r2+1)r!=(r+1)!⋅r−r!⋅(r−1)(r^2+1)r!=(r+1)!\cdot r-r!\cdot(r-1)(r2+1)r!=(r+1)!⋅r−r!⋅(r−1) Check it: r(r+1)!−(r−1)r!=r(r+1)r!−(r−1)r!=[r(r+1)−(r−1)]r!=(r2+1)r!r(r+1)!-(r-1)r!=r(r+1)r!-(r-1)r!=[r(r+1)-(r-1)]r!=(r^2+1)r!r(r+1)!−(r−1)r!=r(r+1)r!−(r−1)r!=[r(r+1)−(r−1)]r!=(r2+1)r! So, (r2+1)r!=r(r+1)!−(r−1)r!(r^2+1)r!=r(r+1)!-(r-1)r!(r2+1)r!=r(r+1)!−(r−1)r!

  2. Now rewrite r(r+1)!r(r+1)!r(r+1)! in a telescoping-friendly form: Let ar=(r−1)(r!)a_r=(r-1)(r!)ar​=(r−1)(r!) Then ar+1=r(r+1)!a_{r+1}=r(r+1)!ar+1​=r(r+1)! Therefore, (r2+1)r!=ar+1−ar(r^2+1)r!=a_{r+1}-a_r(r2+1)r!=ar+1​−ar​

Indeed, ar+1−ar=r(r+1)!−(r−1)r!=(r2+1)r!a_{r+1}-a_r=r(r+1)!-(r-1)r!=(r^2+1)r!ar+1​−ar​=r(r+1)!−(r−1)r!=(r2+1)r!

  1. Hence the sum telescopes: S=∑r=110(ar+1−ar)=a11−a1S=\sum_{r=1}^{10}(a_{r+1}-a_r)=a_{11}-a_1S=∑r=110​(ar+1​−ar​)=a11​−a1​ Now, a11=10⋅11!a_{11}=10\cdot 11!a11​=10⋅11! and a1=(1−1)⋅1!=0a_1=(1-1)\cdot 1!=0a1​=(1−1)⋅1!=0 So, S=10⋅11!S=10\cdot 11!S=10⋅11!

  2. Compare with options:

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

Thus, the correct option is B.

  1. Verification with stored answer: Stored correct answer: B Our derived answer: B So they agree.
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