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

We have to add the infinite series

$$S=(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots,$$

where the numbers satisfy $$|x|\lt 1,\;|y|\lt 1,\;x\neq1.$$ Because the absolute values of both variables are strictly less than unity, every power of $$x$$ and every power of $$y$$ tends to zero, so each individual geometric sub-series we shall encounter is convergent.

First we observe the pattern inside the brackets. In the first bracket the powers of $$x$$ and $$y$$ add up to $$1$$, in the second they add up to $$2$$, in the third they add up to $$3$$, and so on. Concretely, the general $$n^{\text{th}}$$ bracket (where we start counting from $$n=1$$) is

$$x^{n}+x^{\,n-1}y+x^{\,n-2}y^{2}+\ldots+xy^{\,n-1}+y^{n}.$$

That bracket can be written in summation form as

$$\sum_{k=0}^{n}x^{\,n-k}y^{k},$$

because when $$k=0$$ we get $$x^{n}$$, when $$k=1$$ we get $$x^{\,n-1}y$$, and so on all the way to $$k=n$$ where we obtain $$y^{n}.$$

Hence the whole series becomes a double sum:

$$S=\sum_{n=1}^{\infty}\;\sum_{k=0}^{n}x^{\,n-k}y^{k}.$$

Now we change the order of summation. Let us set

$$a=n-k,\qquad b=k.$$

Because $$k$$ runs from $$0$$ to $$n,$$ both $$a$$ and $$b$$ are non-negative integers and they satisfy $$a+b=n.$$ Therefore the pair $$(a,b)$$ runs through all ordered pairs of non-negative integers except the single pair $$(0,0)$$ (that excluded pair would correspond to $$n=0$$, which is not present in the sum). So we can rewrite $$S$$ as

$$S=\sum_{a=0}^{\infty}\;\sum_{b=0}^{\infty}x^{a}y^{b}-1.$$

The “$$-1$$” subtracts the missing pair $$(a,b)=(0,0).$$

Next, we separate the double sum into a product of two independent geometric series. Using the standard formula for the sum of an infinite geometric progression,

$$\sum_{n=0}^{\infty}r^{n}=\frac{1}{1-r}\quad\text{for }|r|\lt 1,$$

we find

$$\sum_{a=0}^{\infty}x^{a}=\frac{1}{1-x},\qquad \sum_{b=0}^{\infty}y^{b}=\frac{1}{1-y}.$$

Hence

$$\sum_{a=0}^{\infty}\;\sum_{b=0}^{\infty}x^{a}y^{b} =\left(\sum_{a=0}^{\infty}x^{a}\right) \left(\sum_{b=0}^{\infty}y^{b}\right) =\frac{1}{(1-x)(1-y)}.$$

Substituting this back, we obtain

$$S=\frac{1}{(1-x)(1-y)}-1.$$

Now we combine the two terms on the right into a single fraction. First place them over a common denominator:

$$S=\frac{1}{(1-x)(1-y)} -\frac{(1-x)(1-y)}{(1-x)(1-y)}.$$

Performing the subtraction in the numerator gives

$$S=\frac{1-(1-x)(1-y)}{(1-x)(1-y)}.$$

We expand the product in the numerator:

$$1-(1-x)(1-y)=1-\bigl[1-x-y+xy\bigr] =1-1+x+y-xy =x+y-xy.$$

Therefore

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

This matches Option C in the given list.

Hence, the correct answer is Option 3.