Question

If S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ... n terms then prove that (x – y)S_n = `[(x^2(x^"n" - 1))/(x - 1) - (y^2(y^"n" - 1))/(y - 1)]`

Answer 1

S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + … n terms

⇒ x.S_n = (x + y)x + (x² + xy + y²)x + (x³ + x²y + xy² + y³)x + ….

⇒ x.S_n = x² +xy + x³ + x²y + y²x + x⁴ + x³y + x²y² + y³x +…..   ...(1)

Multiplying ‘y’ on both sides,

⇒ y.S_n = (x + y)y + (x² + xy + y²)y + (x³ + x²y + xy² + y³)y + ....

⇒ y.S_n = xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + .... .......(2)

(1) – (2)

⇒ x.S_n – y.S_n = (x² + xy + x³ + x²y + y²y + x⁴ + x³y + x²y² + y³x + ...) – (xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + .......)

⇒ (x – y)S_n = (x² + x³ + x⁴ + ......) – (y² + y³ + y⁴ + ......)

= `(x^2(x^"n" - 1))/(x - 1) - (y^2(y^"n" - 1))/(y - 1)`

⇒ (x – y)S_n = `[(x^2(x^"n" - 1))/(x - 1) - (y^2(y^"n" - 1))/(y - 1)]`

Hence proved.