Question

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) + .... is ______.

विकल्प

• `(x + y + xy)/((1 - x)(1 - y))`

• `(x + y - xy)/((1 - x)(1 - y))`

• `(x + y + xy)/((1 + x)(1 + y))`

• `(x + y - xy)/((1 + x)(1 + y))`

Answer 1

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) + .... is `underlinebb((x + y - xy)/((1 - x)(1 - y)))`.

Explanation:

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

Multiply and divide by (x – y)

= `1/((x - y)){(x - y)(x + y) + (x - y)(x^2 + xy + y^2) + (x - y)(x^3 + x^2y + xy^2 + y^3 + ...)}`

= `1/((x - y)){x^2 - y^2 + x^3 - y^3 + x^4 - y^4 + ...}`

= `1/(x - y){x^2 + x^3 + ... -(y^2 + y^3 + ...)}`

= `1/(x - y){x^2/(1 - x) - y^2/(1 - y)}`

= `1/(x - y){(x^2(1 - y) - y^2(1 - x))/((1 - x)(1 - y))}`

= `1/(x - y){(x^2 - y^2 - x^2y + xy^2)/((1 - x)(1 - y))}`

= `1/(x - y)({(x - y)(x + y) - xy(x - y)})/((1 - x)(1 - y))`

= `1/(x - y) ((x - y)(x + y - xy))/((1 - x)(1 - y))`

= `(x + y - xy)/((1 - x)(1 - y))`