Key Concepts and Formulas

• Sum to Infinity of a Geometric Progression (GP): For a GP with first term 'a' and common ratio 'r', where $|r| < 1$, the sum to infinity is given by $S_\infty = \frac{a}{1 - r}$.
• Difference of Powers Factorization: The expression $x^n - y^n$ can be factored as $(x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$.
• Algebraic Simplification: Techniques for combining fractions and factoring algebraic expressions are essential for simplifying the final result.

Step-by-Step Solution

Step 1: Analyze the given series and identify a strategy. The given series is $S = (x + y) + (x^2 + xy + y^2) + (x^3 + x^2y + xy^2 + y^3) + \dots$. Each term in the series is a sum of powers of x and y. The $n$-th term appears to be a sum of terms of the form $x^{n-k}y^k$ for $k$ from 0 to $n-1$. We can recognize that each term is a finite geometric series. For example, the second term $x^2 + xy + y^2$ can be seen as a geometric series with first term $x^2$, common ratio $y/x$, and 3 terms, or as a geometric series with first term $y^2$, common ratio $x/y$, and 3 terms. A more useful observation is the relationship with the difference of powers. The $n$-th term is related to $x^n - y^n$. Specifically, $x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \dots + y^{n-1})$. To utilize this, we can multiply each term by $(x-y)$ and then divide the entire sum by $(x-y)$, since $x \neq y$.

$S = \frac{1}{x-y} \left[ (x-y)(x + y) + (x-y)(x^2 + xy + y^2) + (x-y)(x^3 + x^2y + xy^2 + y^3) + \dots \right]$

Step 2: Apply the difference of powers factorization. Using the identity $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots + b^{n-1})$, we can simplify each product in the bracket. The first term inside the bracket is $(x-y)(x+y) = x^2 - y^2$. The second term inside the bracket is $(x-y)(x^2 + xy + y^2) = x^3 - y^3$. The third term inside the bracket is $(x-y)(x^3 + x^2y + xy^2 + y^3) = x^4 - y^4$. Continuing this pattern, the $n$-th term in the original series, which is $x^n + x^{n-1}y + \dots + y^n$, when multiplied by $(x-y)$, becomes $x^{n+1} - y^{n+1}$. So, the series inside the bracket becomes: $(x^2 - y^2) + (x^3 - y^3) + (x^4 - y^4) + \dots$ Therefore, the sum S can be written as: $S = \frac{1}{x-y} \left[ (x^2 - y^2) + (x^3 - y^3) + (x^4 - y^4) + \dots \right]$

Step 3: Separate the series into two distinct geometric progressions. We can rearrange the terms within the bracket to group terms involving x and terms involving y: $S = \frac{1}{x-y} \left[ (x^2 + x^3 + x^4 + \dots) - (y^2 + y^3 + y^4 + \dots) \right]$ Now we have two infinite series. The first series is $x^2 + x^3 + x^4 + \dots$, which is a geometric progression with first term $a_1 = x^2$ and common ratio $r_1 = x$. The second series is $y^2 + y^3 + y^4 + \dots$, which is a geometric progression with first term $a_2 = y^2$ and common ratio $r_2 = y$.

Step 4: Apply the sum to infinity formula for geometric progressions. We are given that $|x| < 1$ and $|y| < 1$. This condition ensures that the common ratios are less than 1 in absolute value, so both infinite geometric series converge. The sum of the first series is $S_1 = \frac{a_1}{1 - r_1} = \frac{x^2}{1 - x}$. The sum of the second series is $S_2 = \frac{a_2}{1 - r_2} = \frac{y^2}{1 - y}$. Substituting these sums back into the expression for S: $S = \frac{1}{x-y} \left[ \frac{x^2}{1 - x} - \frac{y^2}{1 - y} \right]$

Step 5: Simplify the algebraic expression. To simplify the expression in the bracket, we find a common denominator: $\frac{x^2}{1 - x} - \frac{y^2}{1 - y} = \frac{x^2(1 - y) - y^2(1 - x)}{(1 - x)(1 - y)}$ Expand the numerator: $x^2(1 - y) - y^2(1 - x) = x^2 - x^2y - y^2 + xy^2$ Now, rearrange the numerator to facilitate factorization: $x^2 - y^2 - x^2y + xy^2 = (x^2 - y^2) - (x^2y - xy^2)$ Factor out common terms: $(x^2 - y^2) - xy(x - y)$ Further factor $x^2 - y^2$ as $(x-y)(x+y)$: $(x - y)(x + y) - xy(x - y)$ Now, factor out $(x - y)$ from both terms: $(x - y)(x + y - xy)$ Substitute this back into the expression for S: $S = \frac{1}{x-y} \left[ \frac{(x - y)(x + y - xy)}{(1 - x)(1 - y)} \right]$ Since $x \neq y$, we can cancel the $(x - y)$ term from the numerator and the denominator: $S = \frac{x + y - xy}{(1 - x)(1 - y)}$

Common Mistakes & Tips

• Incorrectly applying the difference of powers formula: Ensure that the correct powers of x and y are used when applying the factorization $x^n - y^n$.
• Algebraic errors during simplification: Be meticulous with algebraic manipulations, especially when combining fractions and factoring, as small errors can lead to an incorrect final answer.
• Forgetting the condition $x \neq y$: This condition is crucial for canceling out the $(x-y)$ term. If $x=y$, the original series terms would be different, and the problem would not be well-defined for this approach.

Summary

The problem involves finding the sum to infinity of a series where each term is a sum of powers of x and y. The key to solving this problem is to recognize that each term in the series can be expressed as a difference of powers of x and y by multiplying and dividing by $(x-y)$. This transforms the original series into a form where it can be split into two separate infinite geometric progressions. By applying the formula for the sum to infinity of a GP to each of these series and then simplifying the resulting algebraic expression, we arrive at the final answer.

The final answer is $\boxed{\frac{x + y - xy}{(1 - x)(1 - y)}}$ which corresponds to option (B).