First slide

Binomial theorem for positive integral Index

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},$ the value of $C_{0} + 2 C_{1} + 3 C_{2} + \dots + (n + 1) C_{n}$ is

Moderate

A

$2^{n - 1}$
B

$n (2^{n - 1})$
C

$n (2^{n - 1}) + 2^{n}$
D

$(n + 1) 2^{n}$

Solution

Let $E = C_{0} + 2 C_{1} + 3 C_{2} + \dots + n C_{n - 1} + (n + 1) C_{n} (1)$
Using $C_{r} = C_{n - r}$ , we can rewrite $(1)$ as
$E = (n + 1) C_{0} + n C_{1} + (n - 1) C_{2} + \dots + 2 C_{n - 1} + C_{n} (2)$
Adding $(1)$ and $(2)$ , we get
$\begin{array}{l} 2 E = (n + 2) C_{0} + (n + 2) C_{1} + (n + 2) C_{2} + \dots \\ + (n + 2) C_{n} \\ = (n + 2) \{C_{0} + C_{1} + \dots + C_{n}\} = (n + 2) 2^{n} \\ \Rightarrow E = (n + 2) 2^{n - 1} \end{array}$
Alternative Solution
We have $C_{0} x + C_{1} x^{2} + C_{2} x^{3} + \dots + C_{n} x^{n + 1} = x (1 + x)^{n}$
Differentiating both the sides, we get
$\begin{array}{l} C_{0} + 2 C_{1} x + 3 C_{2} x^{2} + \dots + (n + 1) C_{n} x^{n} \\ = (1 + x)^{n} + n x (1 + x)^{n - 1} \end{array}$ (1)
Putting $x = 1,$ we get
$\begin{matrix} C_{0} + 2 C_{1} + 3 C_{2} + \dots + (n + 1) C_{n} \\ = 2^{n} + n (1) 2^{n - 1} = (n + 2) 2^{n - 1} \end{matrix}$

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