First slide

Binomial theorem for positive integral Index

Question

Coefficient of the constant term in the expansion of $E = {(x^{2 / 3} + 4 x^{1 / 3} + 4)}^{5} {(\frac{1}{x^{1 / 3} - 1} + \frac{1}{x^{2 / 3} + x^{1 / 3} + 1})}^{- 9}$ is

Moderate

A

$74$
B

$98$
C

$148$
D

$168$

Solution

$\begin{array}{l} E = {({(x^{1 / 3} + 2)}^{2})}^{5} {(\frac{x^{2 / 3} + x^{1 / 3} + 1 + x^{1 / 3} - 1}{x - 1})}^{- 9} \\ = {(2 + x^{1 / 3})}^{10} {(\frac{x^{2 / 3} + 2 x^{1 / 3}}{x - 1})}^{- 9} \\ = {(2 + x^{1 / 3})}^{10} \frac{(x - 1)^{9}}{{(x^{1 / 3})}^{9} {(x^{1 / 3} + 2)}^{9}} \\ = (2 + x^{1 / 3}) x^{- 3} (x - 1)^{9} \end{array}$
Constant term in $E$
$= 2 [$ constant term in $x^{- 3} (x - 1)^{9}]$
$= 2 {(- 1)}^{9} [$ coefficient of $x^{3}$ in ${(1 - x)}^{9}]$
$= 2 {(- 1)}^{9} C_{3} (- 1) = 168$

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