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

$$CanWeexpandthefollowing$$$$expression?$$$$(1+x)(1+2x)(1+3x)......(1+nx)$$$$oristhereanyformulaforthis?$$

Commented by Penguin last updated on 19/Feb/18

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

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

$$itsnot$$

Commented by Penguin last updated on 19/Feb/18

$$\mathrm{Why}?$$

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

$$whatis\mathrm{\Gamma }\left(x\right)\stackrel{}{?}$$$$whatis$$

Commented by Penguin last updated on 19/Feb/18

$$\mathrm{Gamma}\mathrm{function}$$$$\mathrm{\Gamma }\left(x\right)=(x-1)!=(x-1)\times (x-2)\times ...\times 3\times 2\times 1$$

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

$$\mathrm{\Gamma }\left(x\right)istheeulerfunctiondefinedby$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dtforx>0.$$

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

Yes

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