Coefficient of x^3 in expansion of (x^2 - x + 1)^{10} (x^2 + 1 )^{15} is equal to

English

Gujarati

Hindi

7.Binomial Theorem

normal

Coefficient of $x^3$ in the expansion of $(x^2 - x + 1)^{10} (x^2 + 1 )^{15}$ is equal to

$-360$

$-240$

$-180$

$60$

Solution

$\left(x^{2}-x+1\right)^{10} \cdot\left(x^{2}+\right)^{15}$

$=\left(1+^{10} \mathrm{C}_{1}\left(\mathrm{x}^{2}-\mathrm{x}\right)+^{10} \mathrm{C}_{2}\left(\mathrm{x}^{2}-\mathrm{x}\right)^{2}+^{10} \mathrm{C}_{3}\left(\mathrm{x}^{2}-\mathrm{x}\right)^{3}\right)$

$\left(^{15} \mathrm{C}_{0}+^{15} \mathrm{C}_{1}, \mathrm{x}^{2}\right)$

$=-2 \cdot^{10} \mathrm{C}_{2} \cdot^{15} \mathrm{C}_{0}-^{10} \mathrm{C}_{1} \cdot^{15} \mathrm{C}_{1}-^{10} \mathrm{C}_{3}=-360$

Standard 11

Mathematics

The term independent of $y$ in the expansion of ${({y^{ – 1/6}} – {y^{1/3}})^9}$ is

medium

The coefficient of $x^4$ in ${\left[ {\frac{x}{2}\,\, – \,\,\frac{3}{{{x^2}}}} \right]^{10}}$ is :

normal

If $\alpha$ and $\beta$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of

$(\mathrm{x}+\sqrt{\mathrm{x}^{2}-1})^{6}+(\mathrm{x}-\sqrt{\mathrm{x}^{2}-1})^{6}$, then

hard

(JEE MAIN-2020)

The term independent of $x$ in the expansion of ${\left( {{x^2} – \frac{{3\sqrt 3 }}{{{x^3}}}} \right)^{10}}$ is

easy

For the natural numbers $m, n$, if $(1-y)^{m}(1+y)^{n}=1+a_{1} y+a_{2} y^{2}+\ldots .+a_{m+n} y^{m+n}$ and $a_{1}=a_{2}$ $=10$, then the value of $(m+n)$ is equal to:

hard

(JEE MAIN-2021)