That's not a very big gap.

Algebra Level 5

Let $A_n = \max \left\{ \dbinom{n}{k} : 0 \le k \le n \right\}$ where $n$ is a positive integer between 1 and 20 (both inclusive). Find the number of such $n$ satisfying

$\large 1.9 \le \dfrac{A_n}{A_{n-1}} \le 2$

The answer is 11.

1 solution

Patrick Corn
Oct 24, 2017

When $n$ is even, $A_n = \binom{n}{n/2}.$ When $n$ is odd, $A_n = \binom{n}{(n-1)/2}.$ Then it's not hard to see that $\frac{A_n}{A_{n-1}} = \begin{cases} 2 &\text{ if } n \text{ is even} \\ \frac{2n}{n+1}&\text{ if } n \text{ is odd} \end{cases}.$ So $\frac{A_n}{A_{n-1}} \le 2$ always, and if $n$ is even it's $\ge 1.9,$ and if $n$ is odd, setting $\frac{2n}{n+1} \ge 1.9$ is equivalent to $n \ge 19.$ So every even number, and $19,$ satisfy the inequality. The answer is $\fbox{11}.$

Can you please elaborate that how is the value of An found?? How do you know the maximum occurs at that point only??

Aaghaz Mahajan - 3 years, 7 months ago

Well, $\frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{n-k}{k+1}.$ This ratio is $>1$ if $k < (n-1)/2,$ and $<1$ if $k > (n-1)/2.$ Now think about what happens to $\binom{n}{k}$ as you increment $k$ by $1$ (either up or down).

Patrick Corn - 3 years, 7 months ago

A brief and rigorous approach! (+1)

Tapas Mazumdar - 3 years, 7 months ago

That's not a very big gap.

The answer is 11.

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