Question

\[ \sum_{r=1}^{15} r^2 \left( \frac{{}^{15}C_r}{{}^{15}C_{r-1}} \right) =\ ? \]

• A. \(560\)
• B. \(680\)
• C. \(840\)
• D. \(1020\)

Hint:

Use the identity \(\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r}\) to simplify such binomial expressions.

Answer 1

The Correct Option is B

Step 1: Use the identity of binomial coefficients \[ \frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \frac{15 - r + 1}{r} = \frac{16 - r}{r} \] Step 2: Plug into the given sum \[ \sum_{r=1}^{15} r^2 . \frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \sum_{r=1}^{15} r^2 . \frac{16 - r}{r} = \sum_{r=1}^{15} r(16 - r) \] \[ = \sum_{r=1}^{15} (16r - r^2) = 16 \sum_{r=1}^{15} r - \sum_{r=1}^{15} r^2 \] Step 3: Use formulae for sums \[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2},
\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \] \[ \sum_{r=1}^{15} r = \frac{15 . 16}{2} = 120,
\sum_{r=1}^{15} r^2 = \frac{15 . 16 . 31}{6} = 1240 \] \[ \Rightarrow 16 . 120 - 1240 = 1920 - 1240 = 680 \] % Final Answer \[ \boxed{680} \] % Tip