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For positive real numbers $a$ and $b,$ let $$a \diamond b = \frac{\sqrt{a^2 + 4ab + b^2 - 2a - 2b + 9}}{ab + 6}.$$Find $$( \dotsb ((2017 \diamond 2016) \diamond 2015) \diamond \dotsb \diamond 2) \diamond 1.$$


I can't find any quick way to do this. Can anyone help? Thanks in advance!

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  • $\begingroup$ There's nothing other than more parentheses in that leading ellipsis, right? $\endgroup$
    – Brian Tung
    Jun 1, 2020 at 21:26
  • $\begingroup$ yes, there are only the opening parentheses of all the *functions* $\endgroup$
    – Mike Smith
    Jun 1, 2020 at 21:33
  • $\begingroup$ Similar to my question PSE: puzzling.stackexchange.com/questions/99236/… $\endgroup$ Jul 19, 2020 at 12:04

1 Answer 1

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Notice for all positive number $a$, we have

$$a \diamond 2 = \frac{\sqrt{a^2 + 8a + 4 - 2a - 4 + 9}}{2a+6} = \frac{\sqrt{a^2+6a+9}}{2a+6} = \frac12$$

The mess at hand equals to $$( (\cdots) \diamond 2) \diamond 1 = \frac12 \diamond 1 = \frac{\sqrt{37}}{13}$$

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  • $\begingroup$ does it follow from the fact the $((a \diamond b) \diamond c) = (a \diamond (b \diamond c))$? $\endgroup$
    – Alex
    Jun 1, 2020 at 22:25
  • 1
    $\begingroup$ @Alex, nope. The first thing I do is check whether the $\diamond$ operation is associative and it isn't. $\endgroup$ Jun 1, 2020 at 22:26
  • $\begingroup$ In your calculation, $a \diamond 2$, what is $a$? All terms between 2017 and 3 or 3? $\endgroup$
    – Alex
    Jun 1, 2020 at 22:33
  • $\begingroup$ @Alex $a$ can be any positive number. if you substitute it by all the terms between 2017 and 3, you get the mess between 2017 and 2. $\endgroup$ Jun 1, 2020 at 22:36
  • $\begingroup$ OK, so in you construct $a=(\ldots)$, i.e. $\diamond$ between all terms between 3 and 2017. Is it guaranteed to be positive? $\endgroup$
    – Alex
    Jun 1, 2020 at 22:47

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