JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 23)

If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 ; x \neq 0$ is $\alpha$, then $|\alpha|$ equals ___________.
Answer
678

Explanation

$\begin{aligned} & \text { Coefficient of } x^{30} \text { in } \frac{(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8}{x^6} \\\\ & \Rightarrow \text { Coefficient of } x^{36} \text { in }(1+x)^6\left(1+x^2\right)^7\left(1-x^3\right)^8 \\\\ & \Rightarrow \text { General term }={ }^6 C_{r_1}{ }^7 C_{r_2}{ }^8 C_{r_3}(-1)^{r_3} x^{r_1+2 r_2+3 r_3} \\\\ & \Rightarrow r_1+2 r_2+3 r_3=36\end{aligned}$

$$ \text { Case-I : } \begin{array}{|c|c|c|} \hline \mathrm{r}_1 & \mathrm{r}_2 & \mathrm{r}_3 \\ \hline 0 & 6 & 8 \\ \hline 2 & 5 & 8 \\ \hline 4 & 4 & 8 \\ \hline 6 & 3 & 8 \\ \hline \end{array} $$

$r_1+2 r_2=12 \quad\left(\right.$ Taking $\left.r_3=8\right)$

$$ \begin{aligned} &\text { Case-II :}\\\\ &\begin{array}{|l|l|l|} \hline r_1 & r_2 & r_3 \\ \hline 1 & 7 & 7 \\ \hline 3 & 6 & 7 \\ \hline 5 & 5 & 7 \\ \hline \end{array} \end{aligned} $$

$r_1+2 r_2=15\left(\right.$ Taking $\left.r_3=7\right)$

$$ \begin{aligned} &\text { Case-III : }\\\\ &\begin{array}{|l|l|l|} \hline r_1 & r_2 & r_3 \\ \hline 4 & 7 & 6 \\ \hline 6 & 6 & 6 \\ \hline \end{array} \end{aligned} $$

$\begin{aligned} & \text { Coefficient}=7+(15 \times 21)+(15 \times 35)+(35) \\\\ & -(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28) \\\\ & +(7 \times 28)=-678=\alpha \\\\ & |\alpha|=678\end{aligned}$

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