First slide

Binomial theorem for positive integral Index

Question

If $C_{0}, C_{1}, C_{2}, C_{3}, \dots, C_{n}$ denote the binomial coefficients in the binomial expansion of $(1 + x)^{n}$ then $1. C_{1} - 2 \cdot C_{2} + 3 \cdot C_{3} - 4 \cdot C_{4} + \dots + (- 1)^{n - 1} n C_{n} =$

Moderate

A

0
B

$2^{n}$
C

$n 2^{n - 1}$
D

$- n 2^{n - 1}$

Solution

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

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