Question Number 196758 by sonukgindia last updated on 31/Aug/23
Answered by qaz last updated on 31/Aug/23
![1+(1+x)+(1+x+x^2 )+(1+x+x^2 +x^3 )+...(1+x+x^2 +...+x^(n−1) ) =((1−x)/(1−x))+((1−x^2 )/(1−x))+((1−x^3 )/(1−x))+((1−x^4 )/(1−x))+...((1−x^(n−1) )/(1−x)) =(1/(1−x))[n−(x+x^2 +x^3 +...+x^(n−1) )] =(1/(1−x))[n−((x(1−x^n ))/(1−x))]](https://www.tinkutara.com/question/Q196759.png)
$$\mathrm{1}+\left(\mathrm{1}+{x}\right)+\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)+\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)+…\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{{n}−\mathrm{1}} \right) \\ $$$$=\frac{\mathrm{1}−{x}}{\mathrm{1}−{x}}+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}−{x}}+\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{1}−{x}}+\frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}−{x}}+…\frac{\mathrm{1}−{x}^{{n}−\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{x}}\left[{n}−\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…+{x}^{{n}−\mathrm{1}} \right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{x}}\left[{n}−\frac{{x}\left(\mathrm{1}−{x}^{{n}} \right)}{\mathrm{1}−{x}}\right] \\ $$