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Question Number 33147 by Ahmad Hajjaj last updated on 11/Apr/18

If x^m occurs in the expansion of (x + (1/x^2 ))^(2n) , the coefficient of x^m is

$$\mathrm{If}\:\:\:{x}^{{m}} \:\:\mathrm{occurs}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left({x}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}{n}} ,\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{m}} \:\mathrm{is} \\ $$

Commented by prof Abdo imad last updated on 12/Apr/18

we have (x +(1/x^2 ))^(2n) = Σ_(k=0) ^(2n) C_(2n) ^k x^k (x^(−2) )^(2n−k) = Σ_(k=0) ^(2n) C_(2n) ^k x^(k −4n +2k) = Σ_k C_(2n) ^k x^(3k −4n) so if x^m appears in the expansion we get 3k −4n =m⇒ 3k = m +4n ⇒ k =[((m+4n)/3)] and the coefficient is C_(2n) ^([((m+4n)/3)]) .

$${we}\:{have}\:\:\left({x}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}{n}} \:=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:\:\:{C}_{\mathrm{2}{n}} ^{{k}} {x}^{{k}} \:\:\left({x}^{−\mathrm{2}} \right)^{\mathrm{2}{n}−{k}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\:{x}^{{k}\:−\mathrm{4}{n}\:+\mathrm{2}{k}} \:\:=\:\sum_{{k}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:\:{x}^{\mathrm{3}{k}\:−\mathrm{4}{n}} \:{so}\:{if}\:{x}^{{m}} \\ $$$${appears}\:{in}\:{the}\:{expansion}\:\:{we}\:{get}\:\mathrm{3}{k}\:−\mathrm{4}{n}\:={m}\Rightarrow \\ $$$$\mathrm{3}{k}\:=\:{m}\:+\mathrm{4}{n}\:\Rightarrow\:{k}\:=\left[\frac{{m}+\mathrm{4}{n}}{\mathrm{3}}\right]\:{and}\:{the}\:{coefficient} \\ $$$${is}\:\:{C}_{\mathrm{2}{n}} ^{\left[\frac{{m}+\mathrm{4}{n}}{\mathrm{3}}\right]} . \\ $$

Answered by Rio Mike last updated on 11/Apr/18

Σ_(r=1) ^(2n) ^(2n) C_r .x^(2n−r) ((1/x^2 ))^r ^(2n) C_(r .) x^(2n−r) .1^r .x^(−2r) ^(2n) C_(r. ) x^(2n−3n) hence , x^m = x^(2(m)−3(m)) m= 2m −3m m=−m

$$\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\:^{\mathrm{2}{n}} {C}_{{r}} .{x}^{\mathrm{2}{n}−{r}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{r}} \\ $$$$\:\:^{\mathrm{2}{n}} {C}_{{r}\:.} {x}^{\mathrm{2}{n}−{r}} .\mathrm{1}^{{r}} .{x}^{−\mathrm{2}{r}} \\ $$$$\:\:^{\mathrm{2}{n}} {C}_{{r}.\:} {x}^{\mathrm{2}{n}−\mathrm{3}{n}} \\ $$$${hence}\:,\:{x}^{{m}} =\:{x}^{\mathrm{2}\left({m}\right)−\mathrm{3}\left({m}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}=\:\mathrm{2}{m}\:−\mathrm{3}{m} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}=−{m} \\ $$

Answered by MWSuSon last updated on 25/Apr/20

=C_r ^(2n) (x^(−2) )^r x^(2n−r) =C_r ^(2n) x^(−2r) x^(2n−r) −2r+2n−r=m 2n−m=3r ((2n−m)/3)=r replacing r with ((2n−m)/3) we have (((2n)!)/((((2n−m)/3))!(((6n−2n+m)/3))!)) (((2n)!)/((((2n−m)/3))!(((4n+m)/3))!))

$$={C}_{{r}} ^{\mathrm{2}{n}} \left({x}^{−\mathrm{2}} \right)^{{r}} {x}^{\mathrm{2}{n}−{r}} \\ $$$$={C}_{{r}} ^{\mathrm{2}{n}} {x}^{−\mathrm{2}{r}} {x}^{\mathrm{2}{n}−{r}} \\ $$$$−\mathrm{2}{r}+\mathrm{2}{n}−{r}={m} \\ $$$$\mathrm{2}{n}−{m}=\mathrm{3}{r} \\ $$$$\frac{\mathrm{2}{n}−{m}}{\mathrm{3}}={r} \\ $$$${replacing}\:{r}\:{with}\:\frac{\mathrm{2}{n}−{m}}{\mathrm{3}} \\ $$$${we}\:{have}\:\frac{\left(\mathrm{2}{n}\right)!}{\left(\frac{\mathrm{2}{n}−{m}}{\mathrm{3}}\right)!\left(\frac{\mathrm{6}{n}−\mathrm{2}{n}+{m}}{\mathrm{3}}\right)!} \\ $$$$\frac{\left(\mathrm{2}{n}\right)!}{\left(\frac{\mathrm{2}{n}−{m}}{\mathrm{3}}\right)!\left(\frac{\mathrm{4}{n}+{m}}{\mathrm{3}}\right)!} \\ $$

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