Question

If $|x|<1,|y|<1$, and $xy$, then the sum to infinity of the following series $-(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots$ is:
a) $(x+y+x y) /(1-x)(1-y)$
b) $(x+y-x y) /(1-x)(1-y)$
c) $(x+y+x y) /(1+x)(1+y)$
d) $(x+y-x y) /(1+x)(1+y)$

Answer 1

Hint:Here we have to find the sum to infinity of the given series. It describes the total of a collection of unbounded numbers. The series is referred to as an infinite series if it contains infinite terms, and the sum of the first n terms, Sn, is referred to as a partial sum of the specified infinite series. The limit is known as the sum to infinity of the series and the outcome is known as the sum of infinite of series if the partial sum, or the sum of the first n terms, Sn, given a limit as n increases to infinity.

Formula Used:
Sum $= a/ (1-r)$

Complete step by step Solution:
Given expression is
$(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots .0$
$=(1 / x-y)\left\{\left(x^{2}-y^{2}\right)+\left(x^{3}-y^{3}\right)+\left(x^{4}-y^{4}\right)+\ldots . .00\right.$
To calculate the sum of a series that goes on forever, one uses the sum of the infinite geometric series formula. It is also referred to as the total of infinite GP. Although the series has infinite terms, we discover that the sum of a GP converges to a value.
Sum $= a/ (1-r)$
$=\left(x^{2} /(1-x)\right)-\left(y^{2} /(1-y)\right) /(x-y)$
$=\left\{x^{2}(1-y)-y^{2}(1-x)\right\} /(1-x)(1-y)(x-y)$
$=((x+y)-x y)(x-y) /(1-x)(1-y)(x-y)$
$=(x+y-x y) /(1-x)(1-y)$

Hence, the correct option is 2.

Note: This total can only be calculated if a sequence's terms all converge to zero. It still isn't always done even then. A sum to infinity must be defined for a sequence to converge to zero, which is known as a necessary but not sufficient condition. However, not all convergent sequences have a defined sum to infinity.