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profileRachita Aggarwal
asked 2016-10-29 18:00:14 
$(x+y)+(x^2+xy+y^2)+(x^3+x^2y+xy^2+y^3)+\cdots\text{ to }n\text{ terms }$
Find the sum of the above series?
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profileJaved Khan
answered 2016-10-29 18:20:15 
$(x+y)+(x^2+xy+y^2)+(x^3+x^2y+xy^2+y^3)+\cdots\\
=\dfrac{1}{x-y}\left[(x^2-y^2)+(x^3-y^3)+(x^4-y^4)+\cdots\right]\\
=\dfrac{1}{x-y}\left[(x^2+x^3+x^4+\cdots)-(y^2+y^3+y^4+\cdots)\right]\\
=\dfrac{1}{x-y}\left[\dfrac{x^2(x^n-1)}{x-1}-\dfrac{y^2(y^n-1)}{y-1}\right]~~\text{where } x\ne1;y\ne1, x\ne y$
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