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Binomial theorem for positive integral Index

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Question

The value of $50 \sum_{r = 1}^{49} \frac{2 r^{2} - 48 r + 1}{(50 - r) \cdot^{50} C_{r}}$ is _________.

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Solution

$\begin{matrix} \sum_{r = 1}^{49} \frac{2 r^{2} - 48 r + 1}{(50 - r) \cdot^{50} C_{r}} = \sum_{r = 1}^{49} \frac{(r + 1)^{2} - r (50 - r)}{(50 - r) \cdot^{50} C_{r}} \\ = \sum_{r = 1}^{49} (\frac{(r + 1)^{2}}{(50 - r) \cdot^{50} C_{r}} - \frac{r}{^{50} C_{r}}) \\ = \sum_{r = 1}^{49} (\frac{r + 1}{(50 - r) (\frac{^{50} C_{r}}{r + 1})} - \frac{r}{^{50} C_{r}}) = \sum_{r = 1}^{49} (\frac{r + 1}{(50 - r) (\frac{50!}{(50 - r)! r! (r + 1)})} - \frac{r}{50_{c_{r}}}) \\ = \sum_{r = 1}^{49} (\frac{r + 1}{^{50} C_{r + 1}} - \frac{r}{^{50} C_{r}}) \\ = \frac{50}{^{50} C_{50}} - \frac{1}{^{50} C_{1}} \\ = 50 - \frac{1}{50} \end{matrix}$

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