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Permutations and Combinations question

2016 · 9 Apr · Shift 1 · Q30
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Permutations and Combinations question

2016 · 9 Apr · Shift 1 · Q30

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The value of ∑r=115r2(15Cr15Cr−1)\sum\limits_{r = 1}^{15} {{r^2}} \left( {{{{}^{15}{C_r}} \over {{}^{15}{C_{r - 1}}}}} \right)r=1∑15​r2(15Cr−1​15Cr​​) is equal to :
  1. A
    560
  2. B
    680
  3. C
    1240
  4. D
    1085
View written solutionFree

Correct answer: B

  1. We need to evaluate S=∑r=115r2((15r)(15r−1)).S=\sum_{r=1}^{15} r^2\left(\frac{\binom{15}{r}}{\binom{15}{r-1}}\right).S=∑r=115​r2((r−115​)(r15​)​).

  2. Simplify the binomial coefficient ratio:

=\frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(16-r)!}}.$$ Since $(16-r)!=(16-r)(15-r)!$ and $r!=r(r-1)!$, $$\frac{\binom{15}{r}}{\binom{15}{r-1}}= \frac{(r-1)!(16-r)!}{r!(15-r)!} =\frac{16-r}{r}.$$ So, $$S=\sum_{r=1}^{15} r^2\cdot \frac{16-r}{r} =\sum_{r=1}^{15} r(16-r).$$ 3. Expand the summand: $$r(16-r)=16r-r^2.$$ Thus, $$S=\sum_{r=1}^{15}(16r-r^2)=16\sum_{r=1}^{15}r-\sum_{r=1}^{15}r^2.$$ 4. Use standard formulas: $$\sum_{r=1}^{15} r=\frac{15\cdot 16}{2}=120,$$ $$\sum_{r=1}^{15} r^2=\frac{15\cdot 16\cdot 31}{6}=1240.$$ Hence, $$S=16(120)-1240=1920-1240=680.$$ 5. Compare with options: - A: $560$ - B: $680$ ✅ - C: $1240$ - D: $1085$ Therefore, the correct option is **B**.
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