Question Number 161823 by MathsFan last updated on 22/Dec/21
$$\:{Given}\:{that}\:−\mathrm{1}<{x}<\mathrm{1},\:{find}\:{the} \\ $$$$\:{expansion}\:{of}\:\:\frac{\mathrm{3}−\mathrm{2}{x}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{4}+{x}^{\mathrm{2}} \right)}\:{in} \\ $$$$\:{ascending}\:{power}\:{of}\:{x},\:{up}\:{to}\:{and} \\ $$$$\:{including}\:{the}\:{term}\:{in}\:{x}^{\mathrm{3}} \\ $$
Answered by mr W last updated on 23/Dec/21
![=((3(1−((2x)/3)))/(4(1+x)(1+(x^2 /4)))) =(3/4)(1−((2x)/3))(1−x+x^2 −x^3 +x^4 −...)(1−(x^2 /4)+(x^4 /(16))−...) =(3/4)[1+(−(2/3)−1)x+(1−(1/4)+(2/3))x^2 +(−1−(2/3)+(2/(3×4))+(1/4))x^3 +...] =(3/4)[1−(5/3)x+((17)/(12))x^2 −(5/4)x^3 +...] =(3/4)−(5/4)x+((17)/(16))x^2 −((15)/(16))x^3 +...](https://www.tinkutara.com/question/Q161828.png)
$$=\frac{\mathrm{3}\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}}\right)}{\mathrm{4}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +{x}^{\mathrm{4}} −…\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}^{\mathrm{4}} }{\mathrm{16}}−…\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\left[\mathrm{1}+\left(−\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}\right){x}+\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{3}}\right){x}^{\mathrm{2}} +\left(−\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}×\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\right){x}^{\mathrm{3}} +…\right] \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\left[\mathrm{1}−\frac{\mathrm{5}}{\mathrm{3}}{x}+\frac{\mathrm{17}}{\mathrm{12}}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}{x}^{\mathrm{3}} +…\right] \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{5}}{\mathrm{4}}{x}+\frac{\mathrm{17}}{\mathrm{16}}{x}^{\mathrm{2}} −\frac{\mathrm{15}}{\mathrm{16}}{x}^{\mathrm{3}} +… \\ $$
Commented by peter frank last updated on 23/Dec/21
$$\mathrm{thank}\:\mathrm{you} \\ $$