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Question Number 30267 by Nayon.Sm last updated on 19/Feb/18

Can We expand the following  expression?  (1+x)(1+2x)(1+3x)......(1+nx)  or is there any formula for this?

$${Can}\:{We}\:{expand}\:{the}\:{following} \\ $$$${expression}? \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{2}{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)......\left(\mathrm{1}+{nx}\right) \\ $$$${or}\:{is}\:{there}\:{any}\:{formula}\:{for}\:{this}? \\ $$

Commented by Penguin last updated on 19/Feb/18

f(x)=Π_(k=1) ^n (1+kx)  =(1+x)(1+2x)(1+3x)...(1+nx)  =x^n (1+(1/x))...(n+(1/x))  =x^n (1+(1/x))_n      (x)_n ≡((Γ(x+n))/(Γ(x)))=x(x+1)...(x+n−1)     ∴f(x)=x^n Γ((1/x)+n)Γ(x)^(−1)

$${f}\left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+{kx}\right) \\ $$$$=\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{2}{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)...\left(\mathrm{1}+{nx}\right) \\ $$$$={x}^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)...\left({n}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$={x}^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)_{{n}} \\ $$$$\: \\ $$$$\left({x}\right)_{{n}} \equiv\frac{\Gamma\left({x}+{n}\right)}{\Gamma\left({x}\right)}={x}\left({x}+\mathrm{1}\right)...\left({x}+{n}−\mathrm{1}\right) \\ $$$$\: \\ $$$$\therefore{f}\left({x}\right)={x}^{{n}} \Gamma\left(\frac{\mathrm{1}}{{x}}+{n}\right)\Gamma\left({x}\right)^{−\mathrm{1}} \\ $$

Commented by Nayon.Sm last updated on 19/Feb/18

its not

$${its}\:{not} \\ $$

Commented by Penguin last updated on 19/Feb/18

Why?

$$\mathrm{Why}? \\ $$

Commented by Nayon.Sm last updated on 19/Feb/18

what isΓ(x)?^   what is

$${what}\:{is}\Gamma\left({x}\right)\overset{} {?} \\ $$$${what}\:{is}\: \\ $$

Commented by Penguin last updated on 19/Feb/18

Gamma function  Γ(x)=(x−1)!=(x−1)×(x−2)×...×3×2×1

$$\mathrm{Gamma}\:\mathrm{function} \\ $$$$\Gamma\left({x}\right)=\left({x}−\mathrm{1}\right)!=\left({x}−\mathrm{1}\right)×\left({x}−\mathrm{2}\right)×...×\mathrm{3}×\mathrm{2}×\mathrm{1} \\ $$

Commented by abdo imad last updated on 20/Feb/18

Γ(x) is the euler function defined by  Γ(x)=∫_0 ^∞  t^(x−1) e^(−t) dt  for x>0 .

$$\Gamma\left({x}\right)\:{is}\:{the}\:{euler}\:{function}\:{defined}\:{by} \\ $$$$\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}\:\:{for}\:{x}>\mathrm{0}\:. \\ $$

Commented by Nayon.Sm last updated on 10/Mar/18

Yes

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