How to find the maximum value for integers n?

$\dfrac {n(99)(98)(97) ... (101 - n)}{100^n}$

Any input is appreciated.

Thank you.

#Algebra

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Comments

@Mahdi Raza

Yajat Shamji - 10 months ago

Can you please elaborate the question. I don't understand much

Till where do the dots continue. For ex. when n = 2, what is the expression
and what does the question really ask for, maximum value of n? or number of values for n

All I understand is that this value will always be < 1

Mahdi Raza - 10 months ago

I think for integer n,

The value will be maximum at n=1, since the numerator is getting smaller and denominator is getting bigger with bigger powers, we can't use negatives, as it will make our answer negative, and 0 will make 0. $\dfrac{1(2)(3) \cdots (100)}{100} = 99!$

However, I am not sure, because this is my intuition only. It needs proof or disproof.

Vinayak Srivastava - 10 months ago

$n=100$ gives the minimum value.

Hint: Show that the function $f(n)$ is a decreasing function by showing that the expression $\frac {f(n)}{f(n-1) } > 1$ holds true.

Pi Han Goh - 10 months ago

@Barry Leung - I don't understand..........................................................

A Former Brilliant Member - 10 months ago

I have posted the solution to the Pie Pie Pie question which makes you raise out this question. But the solution is not exactly same as your expression.

Pop Wong - 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$