First slide

Binomial theorem for positive integral Index

Question

If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ are binomial coefficients in the expansion of $(1 + x)^{n}$ then the value of
$C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \frac{C_{3}}{4} + \dots + (- 1)^{n} \frac{C_{n}}{n + 1}$ is

Moderate

A

0
B

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

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

$\frac{- 1}{n + 1}$

Solution

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

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