If the coefficient of x^{30} in the expansion | vedclass

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(C) The expression is $\frac{(x+1)^6}{x^6} (1+x^2)^7 (1-x^3)^8 = \frac{1}{x^6} (1+x)^6 (1+x^2)^7 (1-x^3)^8$.We need the coefficient of $x^{36}$ in $(1

Vedclass · Accepted Answer

$678$

Vedclass · Answer

(C) The expression is $\frac{(x+1)^6}{x^6} (1+x^2)^7 (1-x^3)^8 = \frac{1}{x^6} (1+x)^6 (1+x^2)^7 (1-x^3)^8$.We need the coefficient of $x^{36}$ in $(1+x)^6 (1+x^2)^7 (1-x^3)^8$.The general term is given by $\binom{6}{r_1} \binom{7}{r_2} \binom{8}{r_3} (-1)^{r_3} x^{r_1 + 2r_2 + 3r_3}$.We require $r_1 + 2r_2 + 3r_3 = 36$,where $0 \le r_1 \le 6, 0 \le r_2 \le 7, 0 \le r_3 \le 8$.Calculating the sum of coefficients for all valid triplets $(r_1, r_2, r_3)$ yields $\alpha = -678$.Therefore,$|\alpha| = 678$.

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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