If |x| < 1, |y| < 1 and x ≠ y, then the sum to infinity of the following series
(x + y) + (x²+xy+y²) + (x³+x²y + xy²+y³) + ....

Solution

Maths sir · 4 steps

Multiply each bracket by (x - y) and it collapses to xⁿ⁺¹ - yⁿ⁺¹, so you get two plain GPs.

1. Spot the pattern. The nth bracket is xⁿ + xⁿ⁻¹y + … + yⁿ. Multiplied by (x - y), every middle term cancels and only xⁿ⁺¹ - yⁿ⁺¹ is left. Since x ≠ y, the nth term is (xⁿ⁺¹ - yⁿ⁺¹)/(x - y).
2. Two geometric series. Add over n = 1, 2, 3, …. Because |x| < 1 and |y| < 1, both series converge: x² + x³ + … = x²/(1-x), and likewise for y.

   S = 1/(x-y)(x²/(1-x) - y²/(1-y)) = (x²(1-y) - y²(1-x))/(x-y)(1-x)(1-y)

3. Factor the top. The numerator is x² - y² - xy(x - y) = (x - y)(x + y - xy). The factor (x - y) cancels with the one in the denominator.
4. So. S = (x + y - xy)/(1-x)(1-y), which is option (b).

The idea to keep. When a bracket holds every term xᵃ yᵇ with the same total degree, multiply by (x - y). It telescopes to xⁿ⁺¹ - yⁿ⁺¹, and the series splits into two GPs.

Answer from the NTA final answer key. Solution by PrepWiser.