Discussion
If \(\alpha = {}^{n}C_{r}\) and \(\beta = {}^{n}C_{r-1}\), then \(1 + \dfrac{\alpha}{\beta}\) is equal to
Answer: B
\(1 + \frac{\alpha}{\beta} = \) \(\frac{n+1}{r}\).
The ratio of consecutive binomial coefficients is
\[ \begin{aligned} \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} &= \frac{n - r + 1}{r} \\ 1 + \frac{\alpha}{\beta} &= \frac{r + n - r + 1}{r} = \frac{n + 1}{r} \end{aligned} \]
\(\frac{n - r + 1}{r}\) is \(\frac{\alpha}{\beta}\) alone, without the 1 added. \(\frac{n+1}{r-1}\) and \(\frac{n+1}{r+1}\) have the wrong denominator. Equivalently, the result is Pascal's rule \({}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}\) divided by \({}^{n}C_{r-1}\).
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