JEE Main 2024 — Binomial Theorem Question with Solution

Given expansion is ${\left(1+\frac{1}{x}\right)}^6{\left(1+x^2\right)}^7{\left(1-x^3\right)}^8$.

Now, coefficient of $x^{30}$ in $\frac{{\left(x+1\right)}^6{\left(1+x^2\right)}^7{\left(1-x^3\right)}^8}{x^6}$

So, we will find the coefficient of $x^{36}$ in ${\left(1+x\right)}^6{\left(1+x^2\right)}^7{\left(1-x^3\right)}^8$

General term of the given expansion is${}^6{C_{r_1}}^7{C_{r_2}}^8{C}_{r_3}{\left(-1\right)}^{r_3}{x}^{r_1+2r_2+3r_3}$.

$\Rightarrow r_1+2r_2+3r_3=36$

Case-I:

$\Rightarrow r_1+2r_2=12$ (Taking $r_3=8$)

Case II:

$\Rightarrow r_1+2r_2=15$ (Taking $r_3=7$)

Case III:

$\Rightarrow r_1+2r_2=18$ (Taking $r_3=6$)

So, the required coefficient Is given by, $\alpha =7+\left(15\times 21\right)+\left(15\times 35\right)+\left(35\right)-\left(6\times 8\right)-\left(20\times 7\times 8\right)-\left(6\times 21\times 8\right)+\left(15\times 28\right)+\left(7\times 28\right)$

$\Rightarrow \alpha =-678$

$\Rightarrow \left|\alpha \right|=678$