Question Number 158281 by Ari last updated on 01/Nov/21
Commented by Ari last updated on 01/Nov/21
Find the constant term on the decomposition of expression
Commented by cortano last updated on 02/Nov/21
$$\Rightarrow\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{3}} \frac{\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{4}} }{{x}^{\mathrm{4}} } \\ $$$$=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\begin{pmatrix}{\mathrm{3}}\\{{k}}\end{pmatrix}\:\mathrm{2}^{{k}} \left(\mathrm{3}{x}\right)^{\mathrm{3}−{k}} \:.\frac{\underset{{i}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\begin{pmatrix}{\mathrm{4}}\\{{i}}\end{pmatrix}\left(−\mathrm{4}{x}\right)^{\mathrm{4}−{i}} }{{x}^{\mathrm{4}} } \\ $$$${the}\:{constant}\:{we}\:{get}\:{from} \\ $$$${case}\left(\mathrm{1}\right)\begin{cases}{{k}=\mathrm{3}}\\{{i}=\mathrm{0}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}\left(\mathrm{2}^{\mathrm{3}} \right)\left(\mathrm{3}\right)^{\mathrm{0}} ×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{0}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{3}} ×\mathrm{4}^{\mathrm{4}} =\mathrm{2}^{\mathrm{11}} \\ $$$${case}\left(\mathrm{2}\right)\begin{cases}{{k}=\mathrm{1}}\\{{i}=\mathrm{2}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}\left(\mathrm{2}\right)\left(\mathrm{3}\right)^{\mathrm{2}} ×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}×\mathrm{3}^{\mathrm{2}} ×\mathrm{6}×\mathrm{2}^{\mathrm{4}} =\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{6}} \\ $$$${case}\left(\mathrm{3}\right)\begin{cases}{{k}=\mathrm{2}}\\{{i}=\mathrm{1}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}\left(\mathrm{2}^{\mathrm{2}} \right)\left(\mathrm{3}\right)×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{3}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{2}} \left(−\mathrm{2}^{\mathrm{6}} \right)=−\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{10}} \\ $$$${case}\left(\mathrm{4}\right)\begin{cases}{{k}=\mathrm{0}}\\{{i}=\mathrm{3}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}\left(\mathrm{2}^{\mathrm{0}} \right)\left(\mathrm{3}\right)^{\mathrm{3}} ×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{3}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{3}} ×\mathrm{4}×\left(−\mathrm{4}\right)=−\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{4}} \\ $$$$\therefore\:{totally}=\mathrm{2}^{\mathrm{11}} +\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{6}} −\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{10}} −\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{4}} =−\mathrm{7120} \\ $$