Question

If S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ... n terms then prove that (x – y)S_n = ${\displaystyle [\frac{x^2(x^{\text{n}}-1)}{x-1}-\frac{y^2(y^{\text{n}}-1)}{y-1}]}$

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 + ......)

= ${\displaystyle \frac{x^2(x^{\text{n}}-1)}{x-1}-\frac{y^2(y^{\text{n}}-1)}{y-1}}$

⇒ (x – y)S_n = ${\displaystyle [\frac{x^2(x^{\text{n}}-1)}{x-1}-\frac{y^2(y^{\text{n}}-1)}{y-1}]}$

Hence proved.