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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
$676$
$677$
$678$
$679$
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
${ }^6 \mathrm{C}_{\mathrm{r}_1}{ }^7 \mathrm{C}_{\mathrm{r}_2}{ }^8 \mathrm{C}_{\mathrm{r}_3}(-1)^{\mathrm{I}_3} \mathrm{x}^{\mathrm{I}_1+2 \mathrm{r}_2+3 \mathrm{r}_3}$
$\mathrm{r}_1+2 \mathrm{r}_2+3 \mathrm{r}_3=36$
Case-$I$ : $r_1+2 r_2=12\left(\right.$ Taking $\left.r_3=8\right)$
| $r_1$ | $r_2$ | $r_3$ |
| $0$ | $6$ | $8$ |
| $2$ | $5$ | $8$ |
| $4$ | $4$ | $8$ |
| $6$ | $3$ | $8$ |
case-$II$ $r_1+2 r_2=15\left(\right.$ Taking $\left.r_3=7\right)$
| $r_1$ | $r_2$ | $r_3$ |
| $1$ | $7$ | $7$ |
| $3$ | $6$ | $7$ |
| $5$ | $5$ | $7$ |
case-$III$ $r_1+2 r_2=18\left(\right.$ Taking $\left.r_3=6\right)$
| $r_1$ | $r_2$ | $r_3$ |
| $4$ | $7$ | $6$ |
| $6$ | $6$ | $6$ |
$\begin{aligned} & \text { Coeff. }=7+(15 \times 21)+(15 \times 35)+(35) \\ & -(6 \times 8)-(20 \times 7 \times 8)-(6 \times 21 \times 8)+(15 \times 28) \\ & +(7 \times 28)=-678=\alpha \\ & |\alpha|=678\end{aligned}$
