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

Question

Aicharya · Accepted Answer

Given, Expansion is $\left(1+x+x^2\right)^8$ Coefficient of $x^5$ in above Expansion is $\begin{aligned} & \sum \frac{8 !}{n_{1} ! n_{2} ! n_{3} !}(1)^{n_1} \cdot(x)^{n_2}\left(x^2\right)^{n_3} \ & \therefore \quad\left(n_1+n_2+n_3\right)=8 \ & \text {and } \quad n_2+2 n_3=5 \end{aligned}$ $\begin{array}{c|c|c} n_1 & n_2 & n_3 \ \hline 5 & 3 & 0 \ 5 & 1 & 2 \ 4 & 3 & 1 \end{array}$ Coefficient of $\begin{aligned} f^5 & =\frac{8 !}{5 ! 3 ! 0 !}+\frac{8 !}{5 ! 1 ! 2 !}+\frac{8 !}{4 ! 3 ! 1 !} \ & =56+168+280=504 \end{aligned}$ Hence, option (d) is correct.

Find the coefficient of \(x^5\) in \(\left(1+x+x^2\right)^8\).

Solution