First slide

Binomial theorem for positive integral Index

Question

The coefficient of $x^{r} [0 \leq r \leq (n - 1)]$ in the expansion of $(x + 3)^{n - 1} + (x + 3)^{n - 2} (x + 2) + (x + 3)^{n - 3} (x + 2)^{2} + \dots + (x + 2)^{n - 1}$ is

Moderate

A

$^{n} C_{r} (3^{r} - 2^{n})$
B

$^{n} C_{\cdot} (3^{n - r} - 2^{n - r})$
C

$^{n} C_{r} (3^{r} + 2^{n - 1})$
D

None of these

Solution

We have,
$(x + 3)^{n - 1} + (x + 3)^{n - 2} (x + 2) + (x + 3)^{n - 3} (x + 2)^{2} + \dots + (x + 2)^{n - 1}$
$= \frac{(x + 3)^{n} - (x + 2)^{n}}{(x + 3) - (x + 2)}$
$= (x + 3)^{n} - (x + 2)^{n}$
$(∵ \frac{x^{n} - a^{n}}{x - a} = x^{n - 1} + x^{n - 2} a^{1} + x^{n - 3} a^{2} + \dots + a^{n - 1})$
Therefore, the coefficient of x ^r in the given expression
$\begin{aligned} = coefficient of x^{r} in [(x + 3)^{n} - (x + 2)^{n}] \\ =^{n} C_{r} 3^{n - r} -^{n} C_{r} 2^{n - r} \\ =^{n} C_{r} (3^{n - r} - 2^{n - r}) \end{aligned}$

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