Binomial Theorem Question 1
Question 1 - 2024 (01 Feb Shift 1)
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
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Answer (678)
Solution
coeff of $x^{30}$ in $\frac{(x+1)^{6}\left(1+x^{2}\right)^{7}\left(1-x^{3}\right)^{8}}{x^{6}}$ coeff. of $x^{36}$ in $(1+x)^{6}\left(1+x^{2}\right)^{7}\left(1-x^{3}\right)^{8}$
General term
${ }^{6} \mathrm{C}{\mathrm{r}{1}}{ }^{7} \mathrm{C}{\mathrm{r}{2}}{ }^{8} \mathrm{C}{\mathrm{r}{3}}(-1)^{\mathrm{r}{3}} \mathrm{X}^{\mathrm{r}{1}+2 \mathrm{r}{2}+3 \mathrm{r}{3}}$
$\mathrm{r}{1}+2 \mathrm{r}{2}+3 \mathrm{r}_{3}=36$
Case-I :
| $r_{1}$ | $r_{2}$ | $r_{3}$ |
|---|---|---|
| 0 | 6 | 8 |
| 2 | 5 | 8 |
| 4 | 4 | 8 |
| 6 | 3 | 8 |
$r_{1}+2 r_{2}=12\left(\right.$ Taking $\left.r_{3}=8\right)$
Case II :
| $r_{1}$ | $r_{2}$ | $r_{3}$ |
|---|---|---|
| 1 | 7 | 7 |
| 3 | 6 | 7 |
| 5 | 5 | 7 |
$r_{1}+2 r_{2}=15\left(\right.$ Taking $\left.r_{3}=7\right)$
| $r_{1}$ | $r_{2}$ | $r_{3}$ |
|---|---|---|
| 4 | 7 | 6 |
| 6 | 6 | 6 |
Case-III :
$r_{1}+2 r_{2}=18\left(\right.$ Taking $\left.r_{3}=6\right)$
Coeff. $=7+(15 \times 21)+(15 \times 35)+(35)$
$-(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28)$
$+(7 \times 28)=-678=\alpha$
$|\alpha|=678$
