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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

$$\Rightarrow {(2+3x)}^3\frac{{(1-4x)}^4}{x^4}$$$$=\underset{k=0}{\sum ^3}\left(\begin{array}{c}3\\ k\end{array}\right)2^k{\left(3x\right)}^{3-k}.\frac{\underset{i=0}{\sum ^4}\left(\begin{array}{c}4\\ i\end{array}\right){(-4x)}^{4-i}}{x^4}$$$$theconstantwegetfrom$$$$case\left(1\right)\{\begin{array}{l}k=3\\ i=0\end{array}\Rightarrow \left(\begin{array}{c}3\\ 3\end{array}\right)\left(2^3\right){\left(3\right)}^0\times \left(\begin{array}{c}4\\ 0\end{array}\right){(-4)}^4$$$$\Rightarrow 2^3\times 4^4=2^{11}$$$$case\left(2\right)\{\begin{array}{l}k=1\\ i=2\end{array}\Rightarrow \left(\begin{array}{c}3\\ 1\end{array}\right)\left(2\right){\left(3\right)}^2\times \left(\begin{array}{c}4\\ 2\end{array}\right){(-4)}^2$$$$\Rightarrow 2\times 3^2\times 6\times 2^4=3^3\times 2^6$$$$case\left(3\right)\{\begin{array}{l}k=2\\ i=1\end{array}\Rightarrow \left(\begin{array}{c}3\\ 2\end{array}\right)\left(2^2\right)\left(3\right)\times \left(\begin{array}{c}4\\ 3\end{array}\right){(-4)}^3$$$$\Rightarrow 2^2\times 3^2\times 2^2(-2^6)=-3^2\times 2^{10}$$$$case\left(4\right)\{\begin{array}{l}k=0\\ i=3\end{array}\Rightarrow \left(\begin{array}{c}3\\ 0\end{array}\right)\left(2^0\right){\left(3\right)}^3\times \left(\begin{array}{c}4\\ 3\end{array}\right){(-4)}^1$$$$\Rightarrow 3^3\times 4\times (-4)=-3^2\times 2^4$$$$\therefore totally=2^{11}+3^3\times 2^6-3^2\times 2^{10}-3^2\times 2^4=-7120$$

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