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Question Number 204815 by cortano12 last updated on 28/Feb/24

$$\mathrm{Given}{({3\mathrm{p}}^2-\mathrm{p}+{\mathrm{q}}^3)}^{12},\mathrm{find}\mathrm{the}$$$$\mathrm{coefficient}\mathrm{of}{\mathrm{p}}^{10}{\mathrm{q}}^6$$

Answered by TonyCWX08 last updated on 28/Feb/24

$$$$$$\overline{)\begin{array}{ccc}3p^2& -p& q^3\\ 4& 2& 2\\ 3& 4& 2\\ 2& 6& 2\\ 1& 8& 2\end{array}}$$$$$$$$12+9+6+3$$$$=30$$

Answered by mr W last updated on 28/Feb/24

$${(3p^2-p+q^3)}^{12}$$$$=\sum _{a+b+c=12}\left(\begin{array}{c}12\\ a,b,c\end{array}\right){\left(3p^2\right)}^a{(-p)}^b{\left(q^3\right)}^c$$$$3c=6\Rightarrow c=2$$$$a+b=12-2=10$$$$2a+b=10$$$$\Rightarrow a=0,b=10$$$$coef.of{p}^{10}{q}^{6}is$$$$\left(\begin{array}{c}12\\ 0,10,2\end{array}\right)3^0{(-1)}^{10}{1}^2=\frac{12!}{10!2!}=66✓$$

Commented by mr W last updated on 28/Feb/24

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