Question Number 30425 by abdo imad last updated on 22/Feb/18
![decompose inside R[x] F(x)= (x^(2n) /((x^2 +1)^n )) with n from N and n>0.](Q30425.png)
$${decompose}\:{inside}\:{R}\left[{x}\right]\: \\ $$ $${F}\left({x}\right)=\:\:\:\frac{{x}^{\mathrm{2}{n}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\:\:{with}\:{n}\:{from}\:{N}\:{and}\:{n}>\mathrm{0}. \\ $$
Answered by sma3l2996 last updated on 24/Feb/18
$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{1} \\ $$ $${F}=\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} }{{t}^{\mathrm{2}{n}} } \\ $$ $${F}=\left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{{k}} \\ $$ $${F}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)^{{k}} \\ $$ $$ \\ $$