First slide

Binomial theorem for positive integral Index

Question

${Series}^{n} C_{1} + 1 \cdot^{n} C_{2} + 2 \cdot^{n} C_{3} + \dots + n \cdot^{n} C_{n}$ is equal to

Moderate

A

$n 2^{n - 1}$
B

$2^{n - 1}$
C

$n 2^{n}$
D

$2^{3 - 1}$

Solution

$Since, (1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{x}^{2} + \dots +^{n} C_{n} x^{n}$
$Ondifferentiatingw . r . t ., x, weget$
$n (1 + x)^{n - 1} = 0 +^{n} C_{1} + 2 \cdot x \cdot^{n} C_{2} + \dots + n \cdot^{n} C_{n} x^{n - 1}$
Put x=1, we get
$\begin{aligned} n (1 + 1)^{n - 1} =^{n} C_{1} + 2 \cdot^{n} C_{2} + \dots + n \cdot^{n} C_{n} \\ \Rightarrow n 2^{n - 1} =^{n} C_{1} + 2 \cdot^{n} C_{2} + \dots + n \cdot^{n} C_{n} \end{aligned}$

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