Find the coefficient of x^5 in (1+x+x^2)^8. | vedclass

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

Vedclass · Accepted Answer

$504$

Vedclass · Answer

(D) The general term in the expansion of $(1+x+x^2)^8$ is given by $\frac{8!}{n_1! n_2! n_3!} (1)^{n_1} (x)^{n_2} (x^2)^{n_3}$,where $n_1 + n_2 + n_3 = 8$ and $n_2 + 2n_3 = 5$.We find the non-negative integer solutions for $(n_1, n_2, n_3)$:$1$. If $n_3 = 0$,then $n_2 = 5$,so $n_1 = 8 - 5 - 0 = 3$. Coefficient: $\frac{8!}{3! 5! 0!} = 56$.$2$. If $n_3 = 1$,then $n_2 = 3$,so $n_1 = 8 - 3 - 1 = 4$. Coefficient: $\frac{8!}{4! 3! 1!} = 280$.$3$. If $n_3 = 2$,then $n_2 = 1$,so $n_1 = 8 - 1 - 2 = 5$. Coefficient: $\frac{8!}{5! 1! 2!} = 168$.Summing these coefficients: $56 + 280 + 168 = 504$.Thus,the coefficient of $x^5$ is $504$.

Find the coefficient of $x^5$ in $(1+x+x^2)^8$.

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