Select Goal &
City
Select Goal
Explore
Question:
The ninth term of the expansion $\left(3x-\frac {1}{2x}\right)^8$ is
The ninth term of the expansion $\left(3x-\frac {1}{2x}\right)^8$ is
Updated On: Apr 25, 2024
- $\frac {-1}{512x^9}$
- $\frac {1}{512x^9}$
- $\frac {1}{256.x^8}$
- $\frac {-1}{512 x^8}$
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The general term of the expansion $(x + a)^n$ is
$T_{r+1} =^nC_r x^{n-r} a^r$ .
We have $\left(3x - \frac{1}{2x}\right)^{8} $
Here, $ r=8, x =3x, a = \left(- \frac{1}{2x}\right), n=8 $
$\therefore$ Nineth term $T_{9} =^{8}C_{8} \left(3x\right)^{8-8} \left(\frac{-1}{2x}\right)^{8} $
$= \frac{1}{256 .x^{8}}$
$T_{r+1} =^nC_r x^{n-r} a^r$ .
We have $\left(3x - \frac{1}{2x}\right)^{8} $
Here, $ r=8, x =3x, a = \left(- \frac{1}{2x}\right), n=8 $
$\therefore$ Nineth term $T_{9} =^{8}C_{8} \left(3x\right)^{8-8} \left(\frac{-1}{2x}\right)^{8} $
$= \frac{1}{256 .x^{8}}$
Was this answer helpful?
1
0
Top Questions on Binomial theorem
- Let $m$ and $n$ be the coefficients of the seventh and thirteenth terms respectively in the expansion of \[\left( \frac{1}{3}x^{\frac{1}{3}} + \frac{1}{2x^{\frac{2}{3}}} \right)^{18}.\]Then \[\left(\frac{n}{m}\right)^{\frac{1}{3}}\]is:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- If the term independent of \(x\) in the expansion of \[ \left( \sqrt{ax^2} + \frac{1}{2x^3} \right)^{10} \] is 105, then \(a^2\) is equal to:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $\alpha = \sum_{r=0}^n (4r^2 + 2r + 1) \binom{n}{r}$ and $\beta = \left( \sum_{r=0}^n \frac{\binom{n}{r}}{r+1} \right) + \frac{1}{n+1}$. If $140 < \frac{2\alpha}{\beta} < 281$, then the value of $n$ is _____.
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $0 \leq r \leq n$. If \[^nC_{r+1} : ^nC_r : ^nC_{r-1} = 55 : 35 : 21,\]then $2n + 5r$ is equal to:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}$, $x \neq 0$, is $\alpha \times 2^8 \times \sqrt[5]{3}$, then $25\alpha$ is
- JEE Main - 2024
- Mathematics
- Binomial theorem
View More Questions
Questions Asked in KCET exam
- If \(\lim\limits_{x \rightarrow 0} \frac{\sin(2+x)-\sin(2-x)}{x}\)= A cos B, then the values of A and B respectively are
- KCET - 2023
- limits of trigonometric functions
- The Curie temperatures of Cobalt and iron are 1400K and 1000K respectively. At T = 1600K , the ratio of magnetic susceptibility of Cobalt to that of iron is
- KCET - 2023
- Magnetism and matter
- A particle is in uniform circular motion. Related to one complete revolution of the particle, which among the stataments is incorrect ?
- KCET - 2023
- Uniform Circular Motion
- The modulus of the complex number \(\frac{(1+i)^2(1+3i)}{(2-6i)(2-2i)}\) is
- KCET - 2023
- complex numbers
- The energy gap of an LED is 2.4 eV. When the LED is switched ‘ON’, the momentum of the emitted photons is
- KCET - 2023
- Semiconductor electronics: materials, devices and simple circuits
View More Questions
Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.
SUBSCRIBE TO OUR NEWS LETTER
COLLEGE NOTIFICATIONSEXAM NOTIFICATIONSNEWS UPDATES
Download the Collegedunia app on 
© 2024 Collegedunia Web Pvt. Ltd. All Rights Reserved