If the coefficient of $x^{30}$ in the expansion of $\left(1 + \frac{1}{x}\right)^6 (1 + x^2)^7 (1 - x^3)^8$, $x \neq 0$ is $\alpha$, then $|\alpha|$ equals
Step-by-Step Solution
Key Concept: Rewrite as coefficient of $x^{36}$ in $(1+x)^6(1+x^2)^7(1-x^3)^8$ after multiplying by $x^6$. Use general term $\binom{6}{r_1}\binom{7}{r_2}\binom{8}{r_3}(-1)^{r_3} x^{r_1+2r_2+3r_3}$ and enumerate all cases where $r_1+2r_2+3r_3=36$.
Multiply through by $x^6$: find coeff of $x^{36}$ in $(1+x)^6(1+x^2)^7(1-x^3)^8$. General term: $\binom{6}{r_1}\binom{7}{r_2}\binom{8}{r_3}(-1)^{r_3}$ with $r_1+2r_2+3r_3=36$, $0\le r_1\le6$, $0\le r_2\le7$, $0\le r_3\le8$. Enumerate cases:
- $r_3=8$: $r_1+2r_2=12$ → $(r_1,r_2)=(0,6),(2,5),(4,4),(6,3)$
- $r_3=7$: $r_1+2r_2=15$ → $(1,7),(3,6),(5,5)$
- $r_3=6$: $r_1+2r_2=18$ → $(4,7),(6,6)$
Summing with appropriate signs: Coeff $= 7+(15\times21)+(15\times35)+(35)-(6\times8)-(20\times7\times8)-(6\times21\times8)+(15\times28)+(7\times28) = -678 = \alpha$. So $|\alpha|=678$.
Correct Answer: 678