(C) The general term $T_{r+1}$ in the expansion of $(x^4 - x^{-3})^{15}$ is given by:$T_{r+1} = ^{15}C_r (x^4)^{15-r} (-x^{-3})^r$$T_{r+1} = ^{15}C_r

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

(C) The general term $T_{r+1}$ in the expansion of $(x^4 - x^{-3})^{15}$ is given by:$T_{r+1} = ^{15}C_r (x^4)^{15-r} (-x^{-3})^r$$T_{r+1} = ^{15}C_r (-1)^r x^{60-4r} x^{-3r}$$T_{r+1} = ^{15}C_r (-1)^r x^{60-7r}$To find the coefficient of $x^{32}$,we set the exponent of $x$ equal to $32$:$60 - 7r = 32$$7r = 28$$r = 4$Substituting $r = 4$ into the expression,the coefficient is $^{15}C_4 (-1)^4 = ^{15}C_4$.

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

Easy

The coefficient of $x^{32}$ in the expansion of $(x^4 - \frac{1}{x^3})^{15}$ is

A
$^{15}C_5$
B
$^{15}C_6$
C
$^{15}C_4$
D
$^{15}C_7$

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

The natural number $m$,for which the coefficient of $x$ in the binomial expansion of $\left( x^{m} + \frac{1}{x^{2}} \right)^{22}$ is $1540$,is

DifficultJEE MAIN 2020

View Solution

The coefficient of $t^{50}$ in $(1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47})$ is -

Difficult

View Solution

If the coefficients of the second,third,and fourth terms in the expansion of $(1 + x)^{2n}$ are in $A.P.$,then $2n^2 - 9n + 7$ is equal to

Medium

View Solution

If $p$ and $q$ are respectively the coefficients of $x^{-3}$ and $x^{-5}$ in the expansion of $\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{21}, x > 0$,then $\frac{5p}{4q} = $

EasyTS EAMCET 2021

View Solution

If the constant term in the binomial expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is $405$,then $|k|$ equals

MediumJEE MAIN 2020

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 $x^{32}$ in the expansion of $(x^4 - \frac{1}{x^3})^{15}$ is

Similar Questions

The natural number $m$,for which the coefficient of $x$ in the binomial expansion of $\left( x^{m} + \frac{1}{x^{2}} \right)^{22}$ is $1540$,is

The coefficient of $t^{50}$ in $(1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47})$ is -

If the coefficients of the second,third,and fourth terms in the expansion of $(1 + x)^{2n}$ are in $A.P.$,then $2n^2 - 9n + 7$ is equal to

If $p$ and $q$ are respectively the coefficients of $x^{-3}$ and $x^{-5}$ in the expansion of $\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{21}, x > 0$,then $\frac{5p}{4q} = $

If the constant term in the binomial expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is $405$,then $|k|$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module