(D) We have,$\frac{(1-3 x)^2}{(1-2 x)} = (1 - 6x + 9x^2)(1 - 2x)^{-1}$.Using the binomial expansion $(1 - y)^{-1} = 1 + y + y^2 + y^3 + y^4 + \dots$,w

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) We have,$\frac{(1-3 x)^2}{(1-2 x)} = (1 - 6x + 9x^2)(1 - 2x)^{-1}$.Using the binomial expansion $(1 - y)^{-1} = 1 + y + y^2 + y^3 + y^4 + \dots$,we get $(1 - 2x)^{-1} = 1 + 2x + 4x^2 + 8x^3 + 16x^4 + \dots$.So,the expression is $(1 - 6x + 9x^2)(1 + 2x + 4x^2 + 8x^3 + 16x^4 + \dots)$.To find the coefficient of $x^4$,we multiply terms:$1 \times (16x^4) = 16x^4$$-6x \times (8x^3) = -48x^4$$9x^2 \times (4x^2) = 36x^4$Summing these coefficients: $16 - 48 + 36 = 4$.

Class 11 Mathematics Binomial Theorem General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

Medium

The coefficient of $x^4$ in the expansion of $\frac{(1-3 x)^2}{(1-2 x)}$ is equal to

TS EAMCET 2001 All TS EAMCET

AP EAMCET 2001 All AP EAMCET

A
$1$
B
$2$
C
$3$
D
$4$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Binomial Theorem

Subtopic

General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

Previous Year

TS EAMCET 2001 Paper All TS EAMCET years →

Previous Year

AP EAMCET 2001 Paper All AP EAMCET years →

The sum of the coefficients in the expansion of $(x + y)^n$ is $4096$. The greatest coefficient in the expansion is

MediumAIEEE 2002

View Solution

The coefficient of $x^4$ in the expansion of $\frac{1}{(1-x)(1-2x)(1-3x)}$ is

MediumAP EAMCET 2019

View Solution

Let the sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2}\right)^n, x \neq 0, n \in N$,be $376$. Then the coefficient of $x^4$ is $......$

DifficultJEE MAIN 2023

View Solution

The coefficient of $\frac{1}{x}$ in the expansion of $(1 + x)^n (1 + \frac{1}{x})^n$ is

Difficult

View Solution

The sum of all possible values of $n \in N$ such that the coefficients of $x$,$x^2$,and $x^3$ in the expansion of $(1+x^2)^2(1+x)^n$ are in arithmetic progression is:

DifficultJEE MAIN 2026

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The coefficient of $x^4$ in the expansion of $\frac{(1-3 x)^2}{(1-2 x)}$ is equal to

Similar Questions

The sum of the coefficients in the expansion of $(x + y)^n$ is $4096$. The greatest coefficient in the expansion is

The coefficient of $x^4$ in the expansion of $\frac{1}{(1-x)(1-2x)(1-3x)}$ is

Let the sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^2}\right)^n, x \neq 0, n \in N$,be $376$. Then the coefficient of $x^4$ is $......$

The coefficient of $\frac{1}{x}$ in the expansion of $(1 + x)^n (1 + \frac{1}{x})^n$ is

The sum of all possible values of $n \in N$ such that the coefficients of $x$,$x^2$,and $x^3$ in the expansion of $(1+x^2)^2(1+x)^n$ are in arithmetic progression is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module