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Binomial Coefficients

Let a r = r...

Question

Let $a_{r} = r^{2} \frac{{^{50} C}_{r}}{{^{50} C}_{r - 1}}$ and $b$ denote the coefficient of $x^{49}$ in $(x - a_{1}) (x - a_{2}) . . . . (x - a_{50})$ , find $(- \frac{1}{17}) b$

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Solution

$a_{r} = r^{2} \frac{^{50} C_{r}}{^{50} C_{r - 1}} = r^{2} (\frac{50 - r + 1}{r})$

As we know $\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r} a_{r} = r (51 - r) = 51 r - r^{2}$

$b$ is the coefficient of $x^{49}$ in $(x - a_{1}) (x - a_{2}) (x - a_{3}) . . . . . . . . (x - a_{49}) (x - a_{50})$

$∴ b = - (a_{1} + a_{2} + a_{3} + . . . . . . . . . . . . . . a_{49} + a_{50})$

$∴ b = - \sum_{r = 1}^{50} a_{r} = - \sum_{r = 1}^{50} 51 r - r^{2} = - [(\frac{51 \times 50 \times 51}{2}) - (\frac{50 \times 51 \times 101}{6})]$

Note : We know that $\sum n = \frac{n (n + 1)}{2}, \sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$
$∴ b = - (65025 - 42925) \Rightarrow b = - 22100 \frac{- b}{17} = \frac{22100}{17} = 1300$

Hence, the answer is $1300.$

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