If ${(1 + x)}^{n} = \sum_{r = 0}^{n}{C_{r}x^{r}}$, then $\left( 1 + \frac{C_{1}}{C_{0}} \right)\ \left( 1 + \frac{C_{2}}{C_{1}} \right)\ldots\ \left( 1 + \frac{C_{n}}{C_{n - 1}} \right)$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
We have, $\left( 1 + \frac{C_{1}}{C_{0}} \right)\left( 1 + \frac{C_{2}}{C_{1}} \right)\ldots\left( 1 + \frac{C_{n}}{C_{n - 1}} \right)$
$= \left( 1 + \frac{n}{1} \right)\left( 1 + \frac{\frac{n(n - 1)}{2!}}{n} \right)\ldots\left( 1 + \frac{1}{n} \right)$
$= \frac{(1 + n)}{1} \bullet \frac{(1 + n)}{2} \bullet \frac{(1 + n)}{3}\ldots\frac{(1 + n)}{n} = \frac{{(n + 1)}^{n}}{n!}$

Q. If ${(1 + x)}^{n} = \sum_{r = 0}^{n}{C_{r}x^{r}}$, then $\left( 1 + \frac{C_{1}}{C_{0}} \right)\ \left( 1 + \frac{C_{2}}{C_{1}} \right)\ldots\ \left( 1 + \frac{C_{n}}{C_{n - 1}} \right)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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