The coefficient of x^{10} in the expansion of | vedclass

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(D) The general term in the expansion of $(1+x^2-x^3)^8$ is given by the multinomial theorem as $\frac{8!}{n_1! n_2! n_3!} (1)^{n_1} (x^2)^{n_2} (-x^3

Vedclass · Accepted Answer

$476$

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(D) The general term in the expansion of $(1+x^2-x^3)^8$ is given by the multinomial theorem as $\frac{8!}{n_1! n_2! n_3!} (1)^{n_1} (x^2)^{n_2} (-x^3)^{n_3}$,where $n_1 + n_2 + n_3 = 8$.This simplifies to $\frac{8!}{n_1! n_2! n_3!} (-1)^{n_3} x^{2n_2 + 3n_3}$.We need the coefficient of $x^{10}$,so we set $2n_2 + 3n_3 = 10$.Possible non-negative integer solutions for $(n_2, n_3)$ such that $n_2 + n_3 \le 8$ are:$1$) If $n_3 = 0$,then $2n_2 = 10 \implies n_2 = 5$. Then $n_1 = 8 - 5 - 0 = 3$. The term is $\frac{8!}{3! 5! 0!} (-1)^0 = 56$.$2$) If $n_3 = 2$,then $2n_2 = 10 - 6 = 4 \implies n_2 = 2$. Then $n_1 = 8 - 2 - 2 = 4$. The term is $\frac{8!}{4! 2! 2!} (-1)^2 = 420$.$3$) If $n_3 = 4$,then $2n_2 = 10 - 12 = -2$ (not possible).Summing the coefficients: $56 + 420 = 476$.

The coefficient of $x^{10}$ in the expansion of $(1+x^2-x^3)^8$ is

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