If |x|

Question

(A) Given the series $S = (x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ Multiplying and dividing by $(x-y)$: $S = \frac{(x-y)(x+y)+(x-y)(

Vedclass · Accepted Answer

$\frac{x+y-xy}{(1-x)(1-y)}$

Vedclass · Answer

(A) Given the series $S = (x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ Multiplying and dividing by $(x-y)$: $S = \frac{(x-y)(x+y)+(x-y)(x^{2}+xy+y^{2})+(x-y)(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots}{x-y}$ Using the identity $(x-y)(x^n + x^{n-1}y + \ldots + y^n) = x^{n+1} - y^{n+1}$: $S = \frac{(x^{2}-y^{2})+(x^{3}-y^{3})+(x^{4}-y^{4})+\ldots}{x-y}$ $S = \frac{(x^{2}+x^{3}+x^{4}+\ldots) - (y^{2}+y^{3}+y^{4}+\ldots)}{x-y}$ Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$: $S = \frac{\frac{x^{2}}{1-x} - \frac{y^{2}}{1-y}}{x-y}$ $S = \frac{x^{2}(1-y) - y^{2}(1-x)}{(1-x)(1-y)(x-y)}$ $S = \frac{x^{2} - x^{2}y - y^{2} + xy^{2}}{(1-x)(1-y)(x-y)}$ $S = \frac{(x^{2}-y^{2}) - xy(x-y)}{(1-x)(1-y)(x-y)}$ $S = \frac{(x-y)(x+y) - xy(x-y)}{(1-x)(1-y)(x-y)}$ $S = \frac{(x-y)(x+y-xy)}{(1-x)(1-y)(x-y)}$ $S = \frac{x+y-xy}{(1-x)(1-y)}$

If $|x| < 1, |y| < 1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ is:

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