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Question Number 28312 by abdo imad last updated on 23/Jan/18
let give P_n (x)=(x+1)^(2n) +(x+2)^n −1 and Q(x)= x^2 +3x +2 find R(x) /P_n (x)=R(x) Q(x) .
$${let}\:{give}\:\:{P}_{{n}} \left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} \:+\left({x}+\mathrm{2}\right)^{{n}} −\mathrm{1}\:{and} \\ $$$${Q}\left({x}\right)=\:{x}^{\mathrm{2}} \:+\mathrm{3}{x}\:+\mathrm{2}\:\:{find}\:{R}\left({x}\right)\:/{P}_{{n}} \left({x}\right)={R}\left({x}\right)\:{Q}\left({x}\right)\:. \\ $$
Answered by sma3l2996 last updated on 24/Jan/18
P(x)=R(x)Q(x) so R(x)=((P(x))/(Q(x))) R(x)=(((x+1)^(2n) )/(x^2 +3x+2))+(((x+2)^n )/(x^2 +3x+2))−(1/(x^2 +3x+2)) we have x^2 +3x+2=(x+1)(x+2) so R(x)=(((x+1)^(2n−1) )/(x+2))+(((x+2)^(n−1) )/(x+1))−(1/((x+1)(x+2))) let′s put t=x+1 so x+2=t+1 so t^(2n−1) =(t^(2n−2) −t^(2n−3) +t^(2n−4) −...+1)(t+1)−1 so (x+1)^(2n−1) =(Σ_(k=2) ^(2n) (−1)^k (x+1)^(2n−k) )(x+2)−1 (((x+1)^(2n−1) )/(x+2))=Σ_(k=2) ^(2n) (−1)^k (x+1)^(2n−k) −(1/(x+2)) samething for (((x+2)^(n−1) )/(x+1))=Σ_(k=2) ^n (x+2)^(n−k) +(1/(x+1)) so R(x)=Σ_(k=2) ^(2n) (−1)^k (x+1)^(2n−k) +Σ_(k=2) ^n (x+2)^(n−k) +(1/(x+1))−(1/(x+2))−(1/((x+1)(x+2))) (1/((x+2)(x+1)))=(1/(x+1))−(1/(x+2)) so R(x)=Σ_(k=2) ^n (x+2)^(n−k) +Σ_(k=2) ^(2n) (−1)^k (x+1)^(2n−k)
$${P}\left({x}\right)={R}\left({x}\right){Q}\left({x}\right)\:\:{so}\:\:{R}\left({x}\right)=\frac{{P}\left({x}\right)}{{Q}\left({x}\right)} \\ $$$${R}\left({x}\right)=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} }{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}+\frac{\left({x}+\mathrm{2}\right)^{{n}} }{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}} \\ $$$${we}\:{have}\:\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right) \\ $$$${so}\:{R}\left({x}\right)=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{1}} }{{x}+\mathrm{2}}+\frac{\left({x}+\mathrm{2}\right)^{{n}−\mathrm{1}} }{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$${let}'{s}\:{put}\:\:\:{t}={x}+\mathrm{1}\:\:\:{so}\:\:{x}+\mathrm{2}={t}+\mathrm{1} \\ $$$${so}\:\:{t}^{\mathrm{2}{n}−\mathrm{1}} =\left({t}^{\mathrm{2}{n}−\mathrm{2}} −{t}^{\mathrm{2}{n}−\mathrm{3}} +{t}^{\mathrm{2}{n}−\mathrm{4}} −…+\mathrm{1}\right)\left({t}+\mathrm{1}\right)−\mathrm{1} \\ $$$${so}\:\:\:\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{1}} =\left(\underset{{k}=\mathrm{2}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \left({x}+\mathrm{1}\right)^{\mathrm{2}{n}−{k}} \right)\left({x}+\mathrm{2}\right)−\mathrm{1} \\ $$$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}−\mathrm{1}} }{{x}+\mathrm{2}}=\underset{{k}=\mathrm{2}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \left({x}+\mathrm{1}\right)^{\mathrm{2}{n}−{k}} −\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$${samething}\:\:\:{for}\:\:\frac{\left({x}+\mathrm{2}\right)^{{n}−\mathrm{1}} }{{x}+\mathrm{1}}=\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\left({x}+\mathrm{2}\right)^{{n}−{k}} +\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${so}\:\:{R}\left({x}\right)=\underset{{k}=\mathrm{2}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \left({x}+\mathrm{1}\right)^{\mathrm{2}{n}−{k}} +\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\left({x}+\mathrm{2}\right)^{{n}−{k}} +\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{2}}−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{2}} \\ $$$${so}\:\: \\ $$$${R}\left({x}\right)=\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\left({x}+\mathrm{2}\right)^{{n}−{k}} +\underset{{k}=\mathrm{2}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \left({x}+\mathrm{1}\right)^{\mathrm{2}{n}−{k}} \\ $$

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