If the Coefficient of x 30 in the expansion of (1+1/x ) 6 (1+x 2 ) 7 (1-x 3 ) 8 ; x neq 0 is α , then α equals .

Mathematics · Binomial Theorem

JEE Main 2024 — 1 February, Shift 1 — Question 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 $\qquad$ .

Answer: 678

Numerical answer — enter this value.

Step-by-step solution

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

coeff. of $x^{36}$ in $(1+x)^{6}\left(1+x^{2}\right)^{7}\left(1-x^{3}\right)^{8}$

{General term} ${ }^{6} \mathrm{C}_{\mathrm{r}_{1}}{ }^{7} \mathrm{C}_{\mathrm{r}_{2}}{ }^{8} \mathrm{C}_{\mathrm{r}_{3}}(-1)^{\mathrm{r}_{3}} \mathrm{x}^{\mathrm{r}_{1}+2 \mathrm{r}_{2}+3 \mathrm{r}_{3}}$

$r_{1}+2 r_{2}+3 r_{3}=36$

Coeff. $=7+(15 \times 21)+(15 \times 35)+(35)$ $-(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28)$$+(7 \times 28)=-678=\alpha$

$|\alpha|=678$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients