Verified Solution • 2025–26 Session moderate

QJEEM16Ma47da: JEE Main 2016 Mathematics - Class 11 JEEM16Ma47da Solution

The value of ∑r=115r2(15Cr15Cr−1)\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right) is equal to :

Multiple Choice Options

  • A 560
  • B 680
  • C 1240
  • D 1085

Step-by-Step Verified Solution

We know,

nCrnCr−1=n−r+1r{{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n - r + 1} \over r}

∴\therefore     15Cr15Cr−1=15−r+1r=16−rr{{^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}} = {{15 - r + 1} \over r} = {{16 - r} \over r}

∴\therefore      ∑r=115r2(15Cr15Cr−1)\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {^{15}{C_{r - 1}}}}} \right)

== ∑r=115r2(16−rr)\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{16 - r} \over r}} \right)

=∑r=115(16r−r2)= \sum\limits_{r = 1}^{15} {\left( {16r - {r^2}} \right)}

=16∑r=115r−∑r=115r2= 16\sum\limits_{r = 1}^{15} {r - \sum\limits_{r = 1}^{15} {{r^2}} }

=16×15×162−15×16×316= 16 \times {{15 \times 16} \over 2} - {{15 \times 16 \times 31} \over 6}

==  120 ×\times 16 −- 40 ×\times 31

== 680

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