First slide

Binomial theorem for positive integral Index

Question

$\sum_{r = 0}^{n} \frac{{(- 1)}^{r}^{n + 1} C_{r + 1}}{^{r + 2} C_{r + 1}} =$

Moderate

A

$\frac{n + 2}{n + 1}$
B

$\frac{n + 1}{n + 3}$
C

$\frac{n + 1}{n + 2}$
D

$n - 1$

Solution

We have
$\begin{array}{l} \sum_{r = 0}^{n} \frac{(- 1)^{r n + 1} C_{r + 1}}{^{r + 2} C_{r + 1}} = \sum_{r = 0}^{n} (- 1)^{r} \frac{(n + 1)!}{(n - r)! (r + 2)!} = \sum_{r = 0}^{n} \frac{(- 1)^{r} \cdot (n + 2)!}{(n - r)! (r + 2)!} \cdot \frac{1}{n + 2} \\ = \frac{1}{n + 2} \sum_{r = 0}^{n} (- 1)^{r n + 2} C_{r + 2} = \frac{1}{n + 2} [{}^{n + 2}C_{2} - {}^{n + 2}C_{3} + {}^{n + 2}C_{4} + \dots] \\ = \frac{1}{n + 2} [{}^{n + 2}C_{0} - {}^{n + 2}C_{1} + {}^{n + 2}C_{2} + \dots + {}^{n + 2}C_{n + 2} - {}^{n + 2}C_{0} + {}^{n + 2}C_{1}] \\ = \frac{1}{n + 2} [0 - 1 + n + 2] = \frac{n + 1}{n + 2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App