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Question Number 174469 by Mathspace last updated on 01/Aug/22

$$f\left(x\right)=(x+1)(x+2)....(x+n)$$$$1)calculate{f}^{{}^{\prime }}\left(x\right)(n⩾1)$$$$2)decomposeF=\frac{1}{f}$$

Answered by Ar Brandon last updated on 01/Aug/22

$$\mathrm{y}=(x+1)(x+2)...(x+n)$$$$\mathrm{lny}=\ln (x+1)+\ln (x+2)+\cdot \cdot \cdot +\ln (x+n)$$$$\frac{1}{\mathrm{y}}\cdot \frac{d\mathrm{y}}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\cdot \cdot \cdot +\frac{1}{x+n}$$$$\Rightarrow \frac{d\mathrm{y}}{dx}=\underset{k=1}{\prod ^n}(x+k)\underset{k=1}{\sum ^n}\frac{1}{x+k}$$$$\frac{1}{(x+1)(x+2)...(x+n)}$$$$=\frac{1}{(n-1)!(x+1)}-\frac{1}{(n-2)!\cdot 1!\cdot (x+2)}+\frac{{(-1)}^{n-1}}{(n-1)!(x+n)}$$$$=\underset{k=1}{\sum ^n}\frac{{(-1)}^{k-1}}{(n-k)!(k-1)!(x+k)}$$

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