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Jonask

\[\color{green} {\text{{Born in 1992 ????}}}\]

  • 11 months ago
  • 11 months ago

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  1. AravindG
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    nah I was born is 15 BC

    • 11 months ago
  2. Jonask
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    lol okay i got a math contest question published in 1992

    • 11 months ago
  3. AravindG
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    Then post it !

    • 11 months ago
  4. Jonask
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    \[\text{Find the largest integer,\not exceeding}\huge \prod_{n=1}^{1992}\frac{ 3n+2 }{ 3n+1 }\]

    • 11 months ago
  5. AravindG
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    On my first attempt it seems there is a pattern that is followed involving a sequence of cancellation .Let me see if its right

    • 11 months ago
  6. AravindG
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    my bad ..The pattern didnt help

    • 11 months ago
  7. Jonask
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    so we have \[\frac{ 5 }{ 4 }\frac{ 8 }{ 7 }\frac{ 11 }{ 10 }...\frac{ 3(1992) +2}{3(1992)+1 }\]

    • 11 months ago
  8. Jonask
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    i wonder if this cud help\[\huge \frac{ a_1+a_2+a_3...+a_n }{ n }\le \sqrt[n]{a_1a_2a_3...a_n}\]

    • 11 months ago
  9. saloniiigupta95
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    I think this may help you more , \[(3n+2)/(3n+1) = 1+ 1/(3n+1)\] That would be giving you a sequence directly...

    • 11 months ago
  10. experimentX
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    Using Euler-Maclaurin formula you find that \[ \huge e^{\int_1^{1992} (\log(3x+2) - \log(3x+1))dx + \log(5/4)} = 13\] The answer is 12 though.

    • 11 months ago
  11. experimentX
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    \[ \huge \lfloor e^{\int_1^{1992} (\log(3x+2) - \log(3x+1))dx + \log(5/4)}\rfloor = 13\]

    • 11 months ago
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