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Algebra Level 3

The value of the infinite product below can be expressed as a b \frac{a}{b} for coprime positive integers a a and b . b. Find a + b a+b .

k = 2 ( k 2 1.21 ) ( k 2 0.81 ) \prod_{k=2}^{\infty} \frac{(k^2-1.21)}{(k^2-0.81)}


The answer is 134.

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1 solution

Abhinav Raichur
Jul 10, 2015

Let A k = ( k 2 1.21 ) ( k 2 . 81 ) A_{k} = \frac{(k^2-1.21)}{(k^2-.81)} We are looking for P = k = 2 A k P = \prod_{k=2}^{\infty} A_{k} We will try and convert this into a telescoping product

A k = ( k + 1.1 ) ( k 1.1 ) ( k + 0.9 ) ( k 0.9 ) A_{k} = \frac{(k+1.1)(k-1.1)}{(k+0.9)(k-0.9)} Now let

f ( k ) = k 0.9 f(k) = k - 0.9 g ( k ) = k + 0.9 g(k) = k + 0.9 ........ the product transforms as P = k = 2 f ( k + 2 ) f ( k ) × g ( k 2 ) g ( k ) P = \prod_{k=2}^{\infty} \frac{f(k+2)}{f(k)} \times \frac{g(k-2)}{g(k)} P = ( f ( 4 ) . f ( 5 ) . f ( 6 ) . . . . . f ( 2 ) . f ( 3 ) . f ( 4 ) . . . . ) × ( g ( 0 ) . g ( 1 ) . g ( 2 ) . . . . . g ( 2 ) . g ( 3 ) . g ( 4 ) . . . . . ) P = ( \frac{f(4) . f(5) . f(6) .....}{f(2) . f(3) . f(4) ....} ) \times ( \frac{g(0) . g(1) . g(2) .....}{g(2) . g(3) . g(4) .....} )

All that remains is P = g ( 0 ) . g ( 1 ) f ( 2 ) . f ( 3 ) P = \frac{g(0) . g(1)}{f(2) . f(3)} P = 0.9 × 1.9 1.1 × 2.1 P = \frac{0. 9 \times 1.9}{1.1 \times 2.1} P = 57 77 P = \frac{57}{77}

answer is 134 \boxed{134}

Great problem and great solution!

Ano Maly - 5 years, 11 months ago

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@Ano Maly thank you 😊

Abhinav Raichur - 5 years, 11 months ago

The solution is incomplete

sanghamitra ghosh - 3 years ago

U are assuming that all the terms cancel but that's not the way .. Uneed to find the general terms, do only some cancellation and apply limits as n tends infinty

sanghamitra ghosh - 3 years ago

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