Write the general term in the expansion of $(x^{2}-yx)^{12}, x \neq 0$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
The general term $T_{r+1}$ in the binomial expansion of $(a+b)^{n}$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.
For the expansion of $(x^{2}-yx)^{12}$,we have $a = x^{2}$,$b = -yx$,and $n = 12$.
Substituting these values into the formula:
$T_{r+1} = {}^{12}C_{r} (x^{2})^{12-r} (-yx)^{r}$
$T_{r+1} = {}^{12}C_{r} (x^{24-2r}) (-1)^{r} y^{r} x^{r}$
$T_{r+1} = (-1)^{r} {}^{12}C_{r} x^{24-2r+r} y^{r}$
$T_{r+1} = (-1)^{r} {}^{12}C_{r} x^{24-r} y^{r}$

Explore More

Similar Questions

In the expansion of $\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x > 0$, if the term independent of $x$ is $(221)k$, then $k$ is equal to:

The constant term in the expansion of $\left(1+\frac{1}{x}\right)^{20}\left(30 x(1+x)^{29}+(1+x)^{30}\right)$ is

If the coefficients of $(2 \alpha+4)$-th and $(\alpha-2)$-th terms in the expansion of $(1+x)^{2018}$ are equal,then $\alpha=$

The numerically greatest term in the expansion of $(5+3x)^6$ when $x=1$ is:

The coefficients of two consecutive terms in the expansion of $(1 + x)^n$ will be equal,if

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