$$\color{green} {\text{{Born in 1992 ????}}}$$

Question

aravindg · Answer

nah I was born is 15 BC

anonymous · Answer

lol okay i got a math contest question published in 1992

aravindg · Answer

Then post it !

anonymous · Answer

$$\text{Find the largest integer,\not exceeding}\huge \prod_{n=1}^{1992}\frac{ 3n+2 }{ 3n+1 }$$

aravindg · Answer

On my first attempt it seems there is a pattern that is followed involving a sequence of cancellation .Let me see if its right

aravindg · Answer

my bad ..The pattern didnt help

anonymous · Answer

so we have $$\frac{ 5 }{ 4 }\frac{ 8 }{ 7 }\frac{ 11 }{ 10 }...\frac{ 3(1992) +2}{3(1992)+1  }$$

anonymous · Answer

i wonder if this cud help$$\huge \frac{ a_1+a_2+a_3...+a_n }{ n }\le \sqrt[n]{a_1a_2a_3...a_n}$$

anonymous · Answer

I think this may help you more , 
$$(3n+2)/(3n+1) = 1+  1/(3n+1)$$
That would be giving you a sequence directly...

experimentx · Answer

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.

experimentx · Answer

$$ \huge \lfloor  e^{\int_1^{1992} (\log(3x+2) - \log(3x+1))dx + \log(5/4)}\rfloor = 13$$

experimentx · Answer