First slide

Binomial theorem for positive integral Index

Question

If ${(1 + x + x^{2} + x^{3})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots + a_{3 n} x^{3 n}$ , then the value of $a_{0} + a_{4} + a_{8} + a_{12} + \dots$ is

Moderate

A

-1
B

0
C

$4^{n - 1}$
D

n

Solution

we have
$\begin{array}{l} a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots + a_{3 n} x^{3 n} = {(1 + x + x^{2} + x^{3})}^{n} \\ \Rightarrow (a_{0} + a_{2} x^{2} + a_{4} x^{4} + \dots) + (a_{1} x + a_{3} x^{3} + \dots) = {(1 + x + x^{2} + x^{3})}^{n} \end{array}$
Putting x = 1, -1, i and -i respectively, we get
$\begin{array}{l} (a_{0} + a_{2} + a_{4} + \dots) + (a_{1} + a_{3} + a_{5} + \dots .) = 4^{n} \\ (a_{0} + a_{2} + a_{4} + \dots) - (a_{1} + a_{3} + a_{5} + \dots .) = 0 \\ (a_{0} - a_{2} + a_{4} - a_{6} + \dots) + i (a_{1} - a_{3} + a_{5} + \dots .) = 0 \\ (a_{0} - a_{2} + a_{4} - a_{6} + \dots) - i (a_{1} - a_{3} + a_{5} + \dots .) = 0 \end{array}$
Adding (i) and (ii), we get
$2 (a_{0} + a_{2} + a_{4} + \dots) = 4^{n}$
Adding (iii) and (iv), we get
$2 (a_{0} - a_{2} + a_{4} - a_{6} + \dots . .) = 0$
Adding (v) and (vi), we get
$4 (a_{0} + a_{4} + a_{8} + \dots .) = 4^{n} \Rightarrow a_{0} + a_{4} + a_{8} + \dots . = 4^{n - 1}$

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