## Jonask 2 years ago $\color{green} {\text{{Born in 1992 ????}}}$

1. AravindG

nah I was born is 15 BC

lol okay i got a math contest question published in 1992

3. AravindG

Then post it !

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

5. AravindG

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

6. AravindG

my bad ..The pattern didnt help

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

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

9. saloniiigupta95

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

10. experimentX

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. experimentX

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

12. experimentX