The sequence $\{a_n\}$ is defined as follows: \begin{align*} a_n = \sqrt{1 + \left(1 + \frac 1n \right)^2} + \sqrt{1 + \left(1 - \frac 1n \right)^2}. \end{align*} What is the value of the expression given below? \begin{align*} \frac{4}{a_1} + \frac{4}{a_2} + \dots + \frac{4}{a_{96}}.\end{align*} $\textbf{(A)}\ \sqrt{18241} \qquad\textbf{(B)}\ \sqrt{18625} - 1 \qquad\textbf{(C)}\ \sqrt{18625} \qquad\textbf{(D)}\ \sqrt{19013} - 1\qquad\textbf{(E)}\ \sqrt{19013}$

Let $x_n=\binom{2n}{n}$ for all $n\in\mathbb{Z}^+$. Prove there exist infinitely many finite sets $A,B$ of positive integers, satisfying $A \cap B = \emptyset $, and \[\frac{{\prod\limits_{i \in A} {{x_i}} }}{{\prod\limits_{j\in B}{{x_j}} }}=2012.\]

Given is a prime number $p$. Prove that the number $$p \cdot (p^2 \cdot \frac{p^{p-1}-1}{p-1})!$$ is divisible by $$\prod_{i=1}^{p}(p^i)!.$$

A sequence is called an [i]arithmetic progression of the first order[/i] if the differences of the successive terms are constant. It is called an [i]arithmetic progression of the second order[/i] if the differences of the successive terms form an arithmetic progression of the first order. In general, for $k\geq 2$, a sequence is called an [i]arithmetic progression of the $k$-th order[/i] if the differences of the successive terms form an arithmetic progression of the $(k-1)$-th order. The numbers \[4,6,13,27,50,84\] are the first six terms of an arithmetic progression of some order. What is its least possible order? Find a formula for the $n$-th term of this progression.

Let $a_0=5/2$ and $a_k=a_{k-1}^2-2$ for $k\ge 1.$ Compute \[\prod_{k=0}^{\infty}\left(1-\frac1{a_k}\right)\] in closed form.

A piecewise linear function is defined by \[f(x) = \begin{cases} x & \text{if } x \in [-1, 1) \\ 2 - x & \text{if } x \in [1, 3)\end{cases}\] and $f(x + 4) = f(x)$ for all real numbers $x.$ The graph of $f(x)$ has the sawtooth pattern depicted below. [color=transparent]Diagram from RandomMath.[/color] [center][img width=45]https://i.ibb.co/JW8jH2Dr/image.png[/img][/center] The parabola $x = 34y^2$ intersects the graph of $f(x)$ at finitely many points. The sum of the $y$-coordinates of these intersection points can be expressed in the form $\tfrac{a + b\sqrt c}d,$ where $a, b, c$ and $d$ are positive integers, $a, b,$ and $d$ has greatest common divisor equal to $1,$ and $c$ is not divisible by the square of any prime. Find $a + b + c + d.$

Let $A=\{(x,y): 0\le x,y < 1\}.$ For $(x,y)\in A,$ let \[S(x,y)=\sum_{\frac12\le\frac mn\le2}x^my^n,\] where the sum ranges over all pairs $(m,n)$ of positive integers satisfying the indicated inequalities. Evaluate \[\lim_{(x,y)\to(1,1),(x,y)\in A}(1-xy^2)(1-x^2y)S(x,y).\]

The sequence of Fibonacci numbers $F_0, F_1, F_2, . . .$ is defined by $F_0 = F_1 = 1 $ and $F_{n+2} = F_n+F_{n+1}$ for all $n > 0$. For example, we have $F_2 = F_0 + F_1 = 2, F_3 = F_1 + F_2 = 3, F_4 = F_2 + F_3 = 5$, and $F_5 = F_3 + F_4 = 8$. The sequence $a_0, a_1, a_2, ...$ is defined by $a_n =\frac{1}{F_nF_{n+2}}$ for all $n \ge 0$. Prove that for all $m \ge 0$ we have: $a_0 + a_1 + a_2 + ... + a_m < 1$.

Let $p$ be an odd prime, and put $N=\frac{1}{4} (p^3 -p) -1.$ The numbers $1,2, \dots, N$ are painted arbitrarily in two colors, red and blue. For any positive integer $n \leqslant N,$ denote $r(n)$ the fraction of integers $\{ 1,2, \dots, n \}$ that are red. Prove that there exists a positive integer $a \in \{ 1,2, \dots, p-1\}$ such that $r(n) \neq a/p$ for all $n = 1,2, \dots , N.$ [I]Netherlands[/i]

Define a positive number sequence sequence $\{a_n\}$ by \[a_{1}=1,(n^2+1)a^2_{n-1}=(n-1)^2a^2_{n}.\]Prove that\[\frac{1}{a^2_1}+\frac{1}{a^2_2}+\cdots +\frac{1}{a^2_n}\le 1+\sqrt{1-\frac{1}{a^2_n}} .\]

Find the largest real constant $a$ such that for all $n \geq 1$ and for all real numbers $x_0, x_1, ... , x_n$ satisfying $0 = x_0 < x_1 < x_2 < \cdots < x_n$ we have \[\frac{1}{x_1-x_0} + \frac{1}{x_2-x_1} + \dots + \frac{1}{x_n-x_{n-1}} \geq a \left( \frac{2}{x_1} + \frac{3}{x_2} + \dots + \frac{n+1}{x_n} \right)\]

Let $n$ be a positive integer and let $a_1, a_2, \ldots, a_n$ be positive reals. Show that $$\sum_{i=1}^{n} \frac{1}{2^i}(\frac{2}{1+a_i})^{2^i} \geq \frac{2}{1+a_1a_2\ldots a_n}-\frac{1}{2^n}.$$

For each integer $n\ge 1,$ compute the smallest possible value of \[\sum_{k=1}^{n}\left\lfloor\frac{a_k}{k}\right\rfloor\] over all permutations $(a_1,\dots,a_n)$ of $\{1,\dots,n\}.$ [i]Proposed by Shahjalal Shohag, Bangladesh[/i]

Prove \[ \frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}. \]