Question

sum of series
$(x + y) + ( {x}^{2} + xy + {y}^{2} ) \: + ( {x}^{2} + {x}^{2} y + {xy}^{2} + {y}^{2}) \: is$
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Answer 1

sum of series (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + .......... is

solution : we know, x² - y² = (x - y)(x + y)

⇒(x + y) = (x² - y)/(x - y) .......(1)

x³ - y³ = (x - y)(x² + xy + y²)

⇒(x² + xy + y²) = (x³ - y³)/(x - y) .......(2)

x⁴ - y⁴ = (x - y)(x + y)(x² + y) = (x - y)(x³ + x²y + xy² + y³)

⇒ (x³ + x²y + xy² + y³) = (x⁴ - y⁴)/(x - y) .....(3)

..... so on

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

= (x² - y²)/(x - y) + (x³ - y³)/(x - y) + (x⁴ - y⁴)/(x - y) + ....

= 1/(x - y) [x² - y² + x³ - y³ + x⁴ - y⁴ + .... ]

= 1/(x - y) [ (x² + x³ + x⁴ + ......) - (y² + y³ + y⁴ + .......) ]

[ we know, a + ar + ar² + ar³ + ... = a/(1 - r)]

= 1/(x - y) [ x²/(1 - x) - y²/(1 - y)]

= 1/(x - y)[ {x²(1 - y) - y²(1 - x)}/(1 - x)(1 - y)]

= 1/(x - y) [{x² - x²y - y² + xy²}/(1 - x)(1 - y)]

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

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

Therefore the sum of given series is (x + y - xy)/(1 - x)(1 - y)