First slide

Binomial theorem for positive integral Index

Question

Coefficient of $x^{4}$ in the expansion of ${(1 + 3 x + \frac{2}{x^{2}})}^{6}$ is:

Moderate

A

$240$
B

$735$
C

$820$
D

Solution

The general term in the expansion of ${(1 + 3 x + \frac{2}{x^{2}})}^{6}$ is
$\frac{6!}{a! b! c!} 1^{a} (3 x)^{b} {(\frac{2}{x^{2}})}^{c}$
where $a, b, c \geq 0$ and $a + b + c = 6$
$= \frac{6! 3^{b} 2^{6 - a - b}}{a! b! (6 - a - b)!} x^{b + 2 a + 2 b - 12}$
for coefficient of $x^{4},$ set $2 a + 3 b - 12 = 4$
Now $2 a + 3 b = 16$ and $a + b + c = 6 \Rightarrow (a, b, c) = (2, 4, 0)$
Thus, required coefficient is $\frac{6! (3^{4}) (2^{0})}{2! 4!} = 735$

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