Find the sum of geometric series (x + y) + (x2 + xy + y2) + (x3 + x2 y + xy2 + y3) + …. to n terms 

(x + y) + (x² + xy + y²) + (x³ + x² y + xy² + y³) + …. to n terms;

Let S_n = (x + y) + (x² + xy + y²) + (x³ + x² y + xy² + y³) + …. to n terms

Let us multiply and divide by (x – y) we get,

S_n = \(\dfrac{ 1}{(x – y) [(x + y) (x – y) + (x^2 + xy + y^2) (x – y)}\) … upto n terms]

(x – y) S_n = (x² – y²) + x³ + x²y + xy² – x²y – xy² – y³..upto n terms

(x – y) S_{n =} (x² + x³ + x⁴+…n terms) – (y² + y³ + y⁴ +…n terms)

By using the formula,

Sum of GP for n terms = \(\dfrac{ a(1 – r^n )}{(1 – r)}\)

We have two G.Ps in above sum, so,

S_n = \( \dfrac{1}{(x-y) {x^2 [(x^n – 1)}{ (x – 1)] – y^2 [(y^n – 1)} (y – 1)]}\)