Hidden Geometric Series

Algebra Level 2

Let x x and y y be real numbers with x < 1 |x|< 1 and y < 1 |y| < 1 such that:

{ x + x y + x y 2 + x y 3 + = 4 y + y x + y x 2 + y x 3 + = 7 \begin{cases} x + xy + xy^2 + xy^3 + \dots = 4 \\ y + yx + yx^2 + yx^3 +\dots = 7 \end{cases}

What is 81 x y 81 xy ?

16 56 28 49 81

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1 solution

Hana Wehbi
Feb 5, 2021

x + x y + x y 2 + x y 3 + = 4 x + xy + xy^2 + xy^3 + \dots = 4 The left side is the sum of an infinite geometric series with first term as x x and common ratio as y y .

Since y < 1 |y| < 1 , the sum exists as S = a 1 1 r = x 1 y = 4 4 4 y = x S = \frac{a_1}{1-r} = \frac{x}{1-y} = 4 \implies 4 - 4y = x ;

Similarly: y + y x + y x 2 + y x 3 + = 7 y + yx + yx^2 + yx^3 +\dots = 7 is another infinite geometric series with the first term as y y and common ratio as x x . Since x < 1 |x|<1 , the sum of the the left side of this series exists and it is equal to 7 7 .

Thus, S = a 1 1 r = y 1 x = 7 7 7 x = y S' = \frac{a'_1}{1-r'} = \frac{y}{1-x} = 7\implies 7-7x = y ;

we have two equations with two unknowns: { 7 7 x = y 4 4 y = x \begin{cases} 7 - 7x = y\\ 4 - 4y = x \end{cases}

( x , y ) = ( 8 9 , 7 9 ) 81 x y = 8 × 7 = 56 \implies (x, y) = (\frac{8}{9}, \frac{7}{9}) \implies 81xy = 8\times 7 = 56

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