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Question-159999




Question Number 159999 by Tawa11 last updated on 23/Nov/21
Answered by Kunal12588 last updated on 23/Nov/21
(1−x^n )=(1−x)(1+x+x^2 +x^3 +x^4 +...+x^(n−1) )  ∫((1−x^(100) )/(1−x))dx=∫(1+x+x^2 +x^4 +...+x^(99) )dx  =x+(1/2)x^2 +(1/3)x^3 +...+(1/(100))x^(100) =Σ_(k=1) ^(100) (x^k /k)
$$\left(\mathrm{1}−{x}^{{n}} \right)=\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +…+{x}^{{n}−\mathrm{1}} \right) \\ $$$$\int\frac{\mathrm{1}−{x}^{\mathrm{100}} }{\mathrm{1}−{x}}{dx}=\int\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +…+{x}^{\mathrm{99}} \right){dx} \\ $$$$={x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +…+\frac{\mathrm{1}}{\mathrm{100}}{x}^{\mathrm{100}} =\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\frac{{x}^{{k}} }{{k}} \\ $$
Commented by Tawa11 last updated on 24/Nov/21
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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