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Question-158281

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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
⇒(2+3x)^3 (((1−4x)^4 )/x^4 ) = Σ_(k=0) ^3 ((3),(k) ) 2^k (3x)^(3−k) .((Σ_(i=0) ^4 ((4),(i) )(−4x)^(4−i) )/x^4 ) the constant we get from case(1) { ((k=3)),((i=0)) :}⇒ ((3),(3) )(2^3 )(3)^0 × ((4),(0) )(−4)^4 ⇒2^3 ×4^4 =2^(11) case(2) { ((k=1)),((i=2)) :}⇒ ((3),(1) )(2)(3)^2 × ((4),(2) )(−4)^2 ⇒2×3^2 ×6×2^4 =3^3 ×2^6 case(3) { ((k=2)),((i=1)) :}⇒ ((3),(2) )(2^2 )(3)× ((4),(3) )(−4)^3 ⇒2^2 ×3^2 ×2^2 (−2^6 )=−3^2 ×2^(10) case(4) { ((k=0)),((i=3)) :}⇒ ((3),(0) )(2^0 )(3)^3 × ((4),(3) )(−4)^1 ⇒3^3 ×4×(−4)=−3^2 ×2^4 ∴ totally=2^(11) +3^3 ×2^6 −3^2 ×2^(10) −3^2 ×2^4 =−7120
$$\Rightarrow\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{3}} \frac{\left(\mathrm{1}−\mathrm{4}{x}\right)^{\mathrm{4}} }{{x}^{\mathrm{4}} } \\ $$$$=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\begin{pmatrix}{\mathrm{3}}\\{{k}}\end{pmatrix}\:\mathrm{2}^{{k}} \left(\mathrm{3}{x}\right)^{\mathrm{3}−{k}} \:.\frac{\underset{{i}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\begin{pmatrix}{\mathrm{4}}\\{{i}}\end{pmatrix}\left(−\mathrm{4}{x}\right)^{\mathrm{4}−{i}} }{{x}^{\mathrm{4}} } \\ $$$${the}\:{constant}\:{we}\:{get}\:{from} \\ $$$${case}\left(\mathrm{1}\right)\begin{cases}{{k}=\mathrm{3}}\\{{i}=\mathrm{0}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}\left(\mathrm{2}^{\mathrm{3}} \right)\left(\mathrm{3}\right)^{\mathrm{0}} ×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{0}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{3}} ×\mathrm{4}^{\mathrm{4}} =\mathrm{2}^{\mathrm{11}} \\ $$$${case}\left(\mathrm{2}\right)\begin{cases}{{k}=\mathrm{1}}\\{{i}=\mathrm{2}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}\left(\mathrm{2}\right)\left(\mathrm{3}\right)^{\mathrm{2}} ×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}×\mathrm{3}^{\mathrm{2}} ×\mathrm{6}×\mathrm{2}^{\mathrm{4}} =\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{6}} \\ $$$${case}\left(\mathrm{3}\right)\begin{cases}{{k}=\mathrm{2}}\\{{i}=\mathrm{1}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}\left(\mathrm{2}^{\mathrm{2}} \right)\left(\mathrm{3}\right)×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{3}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{2}} \left(−\mathrm{2}^{\mathrm{6}} \right)=−\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{10}} \\ $$$${case}\left(\mathrm{4}\right)\begin{cases}{{k}=\mathrm{0}}\\{{i}=\mathrm{3}}\end{cases}\Rightarrow\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\end{pmatrix}\left(\mathrm{2}^{\mathrm{0}} \right)\left(\mathrm{3}\right)^{\mathrm{3}} ×\begin{pmatrix}{\mathrm{4}}\\{\mathrm{3}}\end{pmatrix}\left(−\mathrm{4}\right)^{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{3}} ×\mathrm{4}×\left(−\mathrm{4}\right)=−\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{4}} \\ $$$$\therefore\:{totally}=\mathrm{2}^{\mathrm{11}} +\mathrm{3}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{6}} −\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{10}} −\mathrm{3}^{\mathrm{2}} ×\mathrm{2}^{\mathrm{4}} =−\mathrm{7120} \\ $$

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