Question Number 176384 by Ari last updated on 17/Sep/22
$${Find}\:{the}\:{constant}\:{term}\:{in}\:{the} \\ $$$${expansion}\:{of}\:{the}\:{expression} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{3}} \left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)^{\mathrm{4}} \\ $$
Answered by mr W last updated on 17/Sep/22
$$\mathrm{2}^{\mathrm{3}} ×\left(−\mathrm{4}\right)^{\mathrm{4}} +\mathrm{3}×\mathrm{3}×\mathrm{2}^{\mathrm{2}} ×\mathrm{4}×\left(−\mathrm{4}\right)^{\mathrm{3}} +\mathrm{3}×\mathrm{3}^{\mathrm{2}} ×\mathrm{2}×\mathrm{6}×\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} ×\mathrm{4}×\left(−\mathrm{4}\right) \\ $$$$=−\mathrm{2416}\:\checkmark \\ $$
Commented by Tawa11 last updated on 19/Sep/22
$$\mathrm{Great}\:\mathrm{sir}.\:\mathrm{Please}\:\mathrm{what}\:\mathrm{rule}? \\ $$
Commented by mr W last updated on 19/Sep/22
$${i}\:{applied}\:{no}\:{special}\:{rule}\:{or}\:{formula}.\: \\ $$$${i}\:{just}\:{calculated}\:{manually}\:{all}\:{possible} \\ $$$${terms}\:{one}\:{by}\:{one}. \\ $$$${my}\:{motto}\:{is}:\:{if}\:{a}\:{thing}\:{can}\:{be}\:{done}\:{in} \\ $$$${an}\:{easy}\:{way},\:{i}'{ll}\:\:{try}\:{not}\:{to}\:{do}\:{it}\:{in}\:{a}\: \\ $$$${complicated}\:{way}. \\ $$
Commented by mr W last updated on 19/Sep/22
$$\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{3}} ×\left(−\mathrm{4}\right)^{\mathrm{4}} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{3}×\mathrm{3}×\mathrm{2}^{\mathrm{2}} ×\mathrm{4}×\mathrm{1}×\left(−\mathrm{4}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{3}×\mathrm{3}^{\mathrm{2}} ×\mathrm{2}×\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}!}×\mathrm{1}^{\mathrm{2}} ×\left(−\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\mathrm{2}+\mathrm{3}{x}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right)\left(\frac{\mathrm{1}}{{x}}−\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{3}} ×\mathrm{4}×\mathrm{1}^{\mathrm{3}} ×\left(−\mathrm{4}\right) \\ $$
Commented by Tawa11 last updated on 20/Sep/22
$$\mathrm{Wow},\:\mathrm{great}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$
Answered by cortano1 last updated on 18/Sep/22
![p(x)= (2+3x)^3 ((1/x)−4)^4 p(x)=8.2^8 (1+(3/2)x)^3 (1−(1/(4x)))^4 p(x)=2^(11) (1+(3/2)x)^3 (1−(1/(4x)))^4 =2^(11) [ Σ_(i=0) ^3 ((3),(i) )((3/2)x)^i ][ Σ_(j=0) ^4 ((4),(j) ) (−(1/(4x)))^j ] ⇒i−j=0 ; i=j the constant = 2^(11) (1)+2^(11) ((9/2).(−1))+2^(11) (((27)/4).((3/8))) +2^(11) (((27)/8).(−(1/(16)))) =2^(11) (1−(9/2)+((81)/(32))−((27)/(128))) = −2416](https://www.tinkutara.com/question/Q176386.png)
$$\:\mathrm{p}\left(\mathrm{x}\right)=\:\left(\mathrm{2}+\mathrm{3x}\right)^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{x}}−\mathrm{4}\right)^{\mathrm{4}} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{8}.\mathrm{2}^{\mathrm{8}} \left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}\right)^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4x}}\right)^{\mathrm{4}} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{2}^{\mathrm{11}} \:\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}\right)^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4x}}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{11}} \:\left[\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\begin{pmatrix}{\mathrm{3}}\\{\mathrm{i}}\end{pmatrix}\left(\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}\right)^{\mathrm{i}} \:\right]\left[\:\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\begin{pmatrix}{\mathrm{4}}\\{\mathrm{j}}\end{pmatrix}\:\left(−\frac{\mathrm{1}}{\mathrm{4x}}\right)^{\mathrm{j}} \:\right] \\ $$$$\Rightarrow\mathrm{i}−\mathrm{j}=\mathrm{0}\:;\:\mathrm{i}=\mathrm{j} \\ $$$$\mathrm{the}\:\mathrm{constant} \\ $$$$=\:\mathrm{2}^{\mathrm{11}} \left(\mathrm{1}\right)+\mathrm{2}^{\mathrm{11}} \left(\frac{\mathrm{9}}{\mathrm{2}}.\left(−\mathrm{1}\right)\right)+\mathrm{2}^{\mathrm{11}} \left(\frac{\mathrm{27}}{\mathrm{4}}.\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\right) \\ $$$$\:\:\:\:\:+\mathrm{2}^{\mathrm{11}} \left(\frac{\mathrm{27}}{\mathrm{8}}.\left(−\frac{\mathrm{1}}{\mathrm{16}}\right)\right) \\ $$$$\:\:=\mathrm{2}^{\mathrm{11}} \:\left(\mathrm{1}−\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{81}}{\mathrm{32}}−\frac{\mathrm{27}}{\mathrm{128}}\right) \\ $$$$\:\:=\:−\mathrm{2416} \\ $$
Commented by Tawa11 last updated on 19/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$