Let $\{a_n\}$ be a sequence of positive terms such that $a_{n+1}=a_n+ \frac{n^2}{a_n}$ . Let $b_n =a_n-n$ . (1) Are there infinitely many $n$ such that $b_n \ge 0$ ? (2) Prove that there is a positive number $M$ such that $\sum^{\infty}_{n=3} \frac{b_n}{n+1}<M$.

Let $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and \[ \begin{cases} {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\ x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \] for all $n=1,2,3,\ldots$. Prove that they are converges and find their limits.

Let $a_1,a_2, a_3,...$ be a sequence of real numbers which satisfy the relation $a_{n+1} =\sqrt{a_n^2 + 1}$ Suppose that there exists a positive integer $n_0$ such that $a_{2n_0} = 3a_{n_0}$ . Find the value of $a_{46}$.

Prove that $(2m)!(2n)!$ is a multiple of $m!n!(m+n)!$ for any non-negative integers $m$ and $n$.

A sequence $(a_n)_{-\infty}^{-\infty}$ of zeros and ones is given. It is known that $a_n = 0$ if and only if $a_{n-6} + a_{n-5} +...+ a_{n-1}$ is a multiple of $3$, and not all terms of the sequence are zero. Determine the maximum possible number of zeros among $a_0,a_1,...,a_{97}$.

Evaluate the infinite sum \[\sum_{n \equal{} 0}^\infty \binom{2n}{n}\frac {1}{5^n}.\]

Let $T_k = k - 1$ for $k = 1, 2, 3,4$ and \[T_{2k-1} = T_{2k-2} + 2^{k-2}, T_{2k} = T_{2k-5} + 2^k \qquad (k \geq 3).\] Show that for all $k$, \[1 + T_{2n-1} = \left[ \frac{12}{7}2^{n-1} \right] \quad \text{and} \quad 1 + T_{2n} = \left[ \frac{17}{7}2^{n-1} \right],\] where $[x]$ denotes the greatest integer not exceeding $x.$

Is there an infinite sequence of positive real numbers $x_1,x_2,...,x_n$ satisfying for all $n\ge 1$ the relation $x_{n+2}= \sqrt{x_{n+1}}-\sqrt{x_n}$?

Find the greatest natural number $n$ such there exist natural numbers $x_{1}, x_{2}, \ldots, x_{n}$ and natural $a_{1}< a_{2}< \ldots < a_{n-1}$ satisfying the following equations for $i =1,2,\ldots,n-1$: \[x_{1}x_{2}\ldots x_{n}= 1980 \quad \text{and}\quad x_{i}+\frac{1980}{x_{i}}= a_{i}.\]

Let $p$ be a prime number. Let $\mathbb F_p$ denote the integers modulo $p$, and let $\mathbb F_p[x]$ be the set of polynomials with coefficients in $\mathbb F_p$. Define $\Psi : \mathbb F_p[x] \to \mathbb F_p[x]$ by \[ \Psi\left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n a_i x^{p^i}. \] Prove that for nonzero polynomials $F,G \in \mathbb F_p[x]$, \[ \Psi(\gcd(F,G)) = \gcd(\Psi(F), \Psi(G)). \] Here, a polynomial $Q$ divides $P$ if there exists $R \in \mathbb F_p[x]$ such that $P(x) - Q(x) R(x)$ is the polynomial with all coefficients $0$ (with all addition and multiplication in the coefficients taken modulo $p$), and the gcd of two polynomials is the highest degree polynomial with leading coefficient $1$ which divides both of them. A non-zero polynomial is a polynomial with not all coefficients $0$. As an example of multiplication, $(x+1)(x+2)(x+3) = x^3+x^2+x+1$ in $\mathbb F_5[x]$. [i]Proposed by Mark Sellke[/i]

Sequence {$x_n$} satisfies: $x_1=1$ , ${x_n=\sqrt{x_{n-1}^2+x_{n-1}}+x_{n-1}}$ ( ${n>=2}$ ) Find the general term of {$x_n$}

The sequence $a_0, a_1, a_2, ...$ is defined by $a_{n+1} = a^2_n + (a_n - 1)^2$ for $n \ge 0$. Find all rational numbers $a_0$ for which there exist four distinct indices $k, m, p, q$ such that $a_q - a_p = a_m - a_k$.

An eccentric mathematician has a ladder with $ n$ rungs that he always ascends and descends in the following way: When he ascends, each step he takes covers $ a$ rungs of the ladder, and when he descends, each step he takes covers $ b$ rungs of the ladder, where $ a$ and $ b$ are fixed positive integers. By a sequence of ascending and descending steps he can climb from ground level to the top rung of the ladder and come back down to ground level again. Find, with proof, the minimum value of $ n,$ expressed in terms of $ a$ and $ b.$

A sequence of positive real numbers $a_1, a_2, a_3, ... $ satisfies $a_n = a_{n-1} + a_{n-2}$ for all $n \ge 3$. A sequence $b_1, b_2, b_3, ...$ is defined by equations $b_1 = a_1$ , $b_n = a_n + (b_1 + b_3 + ...+ b_{n-1})$ for even $n > 1$ , $b_n = a_n + (b_2 + b_4 + ... +b_{n-1})$ for odd $n > 1$. Prove that if $n\ge 3$, then $\frac13 < \frac{b_n}{n \cdot a_n} < 1$