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Question Number 66498 by miracle wokama last updated on 16/Aug/19

Commented by MJS last updated on 16/Aug/19

$$\mathrm{too}\mathrm{complicated}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}{f}^{\prime }\left(0\right)=1\mathrm{and}{f}^{\prime }\left(1\right)=100$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$wehavef\left(x\right)=1+x+\frac{x^2}{2}+\frac{x^3}{3}+....+\frac{x^{100}}{100}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=1+x+x^2+....+x^{99}\Rightarrow f^{{}^{\prime }}\left(1\right)=1+1+....+1\left(100\times \right)=100$$$$but{f}^{{}^{\prime }}\left(0\right)=1\Rightarrow f^{{}^{\prime }}\left(1\right)=100f^{{}^{\prime }}\left(0\right)$$

Commented by mathmax by abdo last updated on 16/Aug/19

$$noneedtouseallthoserelations....theQissimpleandeazy....$$

Answered by Cmr 237 last updated on 17/Aug/19

$$f\left(x\right)=\underset{k=1}{\sum ^{100}}\underset{k}{\stackrel{x^k}{-}}+1$$$${\mathrm{f}}^{\left(1\right)}\left(\mathrm{x}\right)=\underset{k=1}{\sum ^{100}}{x}^{k-1}$$$$=\frac{1-x^{100}}{1-x}$$$$f^{\left(1\right)}\left(0\right)=1$$$$f^{\left(1\right)}\left(1\right)=100$$$$d.ou{f}^{\left(1\right)}\left(1\right)=100f^{\left(1\right)}\left(0\right)$$$$$$

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