First slide

Binomial theorem for positive integral Index

Question

Suppose ABC is a triangle and n is a natural number, then sum of the series $S = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r} \cos [n A - (n - r) B]$ is

Difficult

A

${(a + b)}^{n}$
B

${(a - b)}^{n}$
C

$c^{n}$
D

$c^{2 n}$

Solution

$\begin{array}{l} S = \sum_{r = 0}^{n}^{n} C_{r} a^{n - r} b^{r} Re [e^{i (n B - (n - r) A}] \\ = Re \sum_{r = 0}^{n}^{n} C_{r} {(a e^{- i B})}^{n - r} {(b e^{i A})}^{r} \\ = Re {(a e^{- i B} + b e^{i A})}^{n} \\ = Re [a (\cos B - i \sin B) + b (\cos A + i \sin A)]^{n} \\ = Re [(a \cos B + b \cos A) \\ - i (a \sin B - b \sin A)]^{n} \end{array}$
But $a \cos B + b \cos A = c$ [projection formula]
and $a \sin B = b \sin A$ [Law of sines]
$∴ S = Re (c^{n}) = c^{n} .$

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