This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 307

Let $\{u_n\}_{n \ge 1}$ be a sequence of real numbers defined as $u_1 = 1$ and \[ u_{n+1} = u_n + \frac{1}{u_n} \text{ for all $n \ge 1$.}\] Prove that $u_n \le \frac{3\sqrt{n}}{2}$ for all $n$.
(a) Find the numbers $a_0,. . . , a_{100}$, such that $a_0 = 0, a_{100} = 1$ and for all $k = 1,. . . , 99$ : $$a_k = \frac12 a_{k- 1} + \frac12 a_{k+1 }$$ (b) Find the numbers $a_0,. . . , a_{100}$, such that $a_0 = 0, a_{100} = 1$ and for all $k = 1,. . . , 99$ : $$a_k = 1+\frac12 a_{k- 1} + \frac12 a_{k+1 }$$.
Given real numbers $x_i \ (i = 1, 2, \cdots, 4k + 2)$ such that \[\sum_{i=1}^{4k +2} (-1)^{i+1} x_ix_{i+1} = 4m \qquad ( \ x_1=x_{4k+3} \ )\] prove that it is possible to choose numbers $x_{k_{1}}, \cdots, x_{k_{6}}$ such that \[\sum_{i=1}^{6} (-1)^{i} k_i k_{i+1} > m \qquad ( \ x_{k_{1}} = x_{k_{7}} \ )\]
Let $ a, b \in \mathbb{N}$ with $ 1 \leq a \leq b,$ and $ M \equal{} \left[\frac {a \plus{} b}{2} \right].$ Define a function $ f: \mathbb{Z} \mapsto \mathbb{Z}$ by \[ f(n) \equal{} \begin{cases} n \plus{} a, & \text{if } n \leq M, \\ n \minus{} b, & \text{if } n >M. \end{cases} \] Let $ f^1(n) \equal{} f(n),$ $ f_{i \plus{} 1}(n) \equal{} f(f^i(n)),$ $ i \equal{} 1, 2, \ldots$ Find the smallest natural number $ k$ such that $ f^k(0) \equal{} 0.$
Let $$x_0 = 5\,\, ,\, \,\,x_{n+1} = x_n +\frac{1}{x_n}.$$ Prove that $45 < x_{1000} < 45.1$.
A sequence ($a_n$) satisfies the following conditions: (i) For each $m \in N$ it holds that $a_{2^m} = 1/m$. (ii) For each natural $n \ge 2$ it holds that $a_{2n-1}a_{2n} = a_n$. (iii) For all integers $m,n$ with $2m > n \ge 1$ it holds that $a_{2n}a_{2n+1} = a_{2^m+n}$. Determine $a_{2000}$. You may assume that such a sequence exists.
Given that ${a_n}$ and ${b_n}$ are two sequences of integers defined by \begin{align*} a_1=1, a_2=10, a_{n+1}=2a_n+3a_{n-1} & ~~~\text{for }n=2,3,4,\ldots, \\ b_1=1, b_2=8, b_{n+1}=3b_n+4b_{n-1} & ~~~\text{for }n=2,3,4,\ldots. \end{align*} Prove that, besides the number $1$, no two numbers in the sequences are identical.
For a positive integer $x$, define a sequence $a_0, a_1, a_2, . . .$ according to the following rules: $a_0 = 1$, $a_1 = x + 1$ and $$a_{n+2} = xa_{n+1} - a_n$$ for all $n \ge 0$. Prove that there exist infinitely many positive integers x such that this sequence does not contain a prime number.
$(MON 2)$ Given reals $x_0, x_1, \alpha, \beta$, find an expression for the solution of the system \[x_{n+2} -\alpha x_{n+1} -\beta x_n = 0, \qquad n= 0, 1, 2, \ldots\]
For positive integers $ n,$ the numbers $ f(n)$ are defined inductively as follows: $ f(1) \equal{} 1,$ and for every positive integer $ n,$ $ f(n\plus{}1)$ is the greatest integer $ m$ such that there is an arithmetic progression of positive integers $ a_1 < a_2 < \ldots < a_m \equal{} n$ for which \[ f(a_1) \equal{} f(a_2) \equal{} \ldots \equal{} f(a_m).\] Prove that there are positive integers $ a$ and $ b$ such that $ f(an\plus{}b) \equal{} n\plus{}2$ for every positive integer $ n.$
The sequence of $a_n$ is determined by the relation $$a_{n+1}=\frac{k+a_n}{1-a_n}$$ where $k > 0$. It is known that $a_{13} = a_1$. What values can $k$ take?
Let $x_0 = 10^9$, $x_n = \frac{x^2_{n-1}+2}{2x_{n-1}}$ for $n > 0$. Prove that $0 < x_{36} - \sqrt2 < 10^{-9}$.
Let $n \ge 2$ be a natural number and suppose that positive numbers $a_0,a_1,...,a_n$ satisfy the equality $(a_{k-1}+a_{k})(a_{k}+a_{k+1})=a_{k-1}-a_{k+1}$ for each $k =1,2,...,n -1$. Prove that $a_n< \frac{1}{n-1}$
Define a sequence $\langle f(n)\rangle^{\infty}_{n=1}$ of positive integers by $f(1) = 1$ and \[f(n) = \begin{cases} f(n-1) - n & \text{ if } f(n-1) > n;\\ f(n-1) + n & \text{ if } f(n-1) \leq n, \end{cases}\] for $n \geq 2.$ Let $S = \{n \in \mathbb{N} \;\mid\; f(n) = 1993\}.$ [b](i)[/b] Prove that $S$ is an infinite set. [b](ii)[/b] Find the least positive integer in $S.$ [b](iii)[/b] If all the elements of $S$ are written in ascending order as \[ n_1 < n_2 < n_3 < \ldots , \] show that \[ \lim_{i\rightarrow\infty} \frac{n_{i+1}}{n_i} = 3. \]
Define the numbers $a_0, a_1, \ldots, a_n$ in the following way: \[ a_0 = \frac{1}{2}, \quad a_{k+1} = a_k + \frac{a^2_k}{n} \quad (n > 1, k = 0,1, \ldots, n-1). \] Prove that \[ 1 - \frac{1}{n} < a_n < 1.\]
Let $P_n$ denote the $n$-th Pell number defined by $P_{n+1}=2P_n+P_{n-1}$, $P_0=0$, $P_1=1$. Furthermore, let $T_n$ denote the $n$-th triangular number, that is $T_n=\binom{n+1}2$. Show that $$\sum_{n=0}^\infty4T_n\cdot\frac{P_n}{3^{n+2}}=P_3+P_4$$ [i]Proposed by Ángel Plaza[/i]
It is known that the sequence $\{a_n\}$ satisfies $a_1=2$, $a_n=2^{2n}a_{n-1}+n\cdot 2^{n^2}$, $(n \ge 2)$, find the general term of $a_n$.
Let $n$ be a natural number. We define sequences $\langle a_i\rangle$ and $\langle b_i\rangle$ of integers as follows. We let $a_0=1$ and $b_0=n$. For $i>0$, we let $$\left( a_i,b_i\right)=\begin{cases} \left(2a_{i-1}+1,b_{i-1}-a_{i-1}-1\right) & \text{if } a_{i-1}<b_{i-1},\\ \left( a_{i-1}-b_{i-1}-1,2b_{i-1}+1\right) & \text{if } a_{i-1}>b_{i-1},\\ \left(a_{i-1},b_{i-1}\right) & \text{if } a_{i-1}=b_{i-1}.\end{cases}$$ Given that $a_k=b_k$ for some natural number $k$, prove that $n+3$ is a power of two.
Let $N_0$ denote the set of nonnegative integers and $Z$ the set of all integers. Let a function $f : N_0 \times Z \to Z$ satisfy the conditions (i) $f(0, 0) = 1$, $f(0, 1) = 1$ (ii) for all $k, k \ne 0, k \ne 1$, $f(0, k) = 0$ and (iii) for all $n \ge 1$ and $k, f(n, k) = f(n -1, k) + f(n- 1, k - 2n)$. Find the value of $$\sum_{k=0}^{2009 \choose 2} f(2008, k)$$
The sequences $(x_n)$ and $(y_n)$ are defined as follows: $$ x_{n+1} = \frac{x_n+2}{x_n+1},\quad y_{n+1}=\frac{y_n^2+2}{2y_n} \quad \text{ for } n= 0,1,2,\ldots.$$ Prove that for every integer $ n\geq 0 $ the equality $ y_n = x_{2^n-1} $ holds.
Let $\{x_n\}$ be a sequence defined by $x_1 = 2$ and $x_{n+1} = x_n^2 - x_n + 1$ for $n \ge 1$. Prove that $$1 -\frac{1}{2^{2^{n-1}}} < \frac{1}{x_1}+\frac{1}{x_2}+ ... +\frac{1}{x_n}< 1 -\frac{1}{2^{2^n}}$$ for all $n$
Solve the equation ("$2$" encounters $1985$ times): $$\dfrac{x}{2+ \dfrac{x}{2+\dfrac{x}{2+... \dfrac{x}{2+\sqrt {1+x}}}}}=1$$
Define the sequence $A_1, A_2, A_3, \dots$ by $A_1 = 1$ and for $n=1,2,3,\dots$ $$A_{n+1}=\frac{A_n+2}{A_n +1}.$$ Define the sequences $B_1, B_2, B_3,\dots$ by $B_1=1$ and for $n=1,2,3,\dots$ $$B_{n+1}=\frac{B_n^2 +2}{2B_n}.$$ Prove that $B_{n+1}=A_{2^n}$ for all non-negative integers $n$.
The sequence $f_1, f_2, \cdots, f_n, \cdots $ of functions is defined for $x > 0$ recursively by \[f_1(x)=x , \quad f_{n+1}(x) = f_n(x) \left(f_n(x) + \frac 1n \right)\] Prove that there exists one and only one positive number $a$ such that $0 < f_n(a) < f_{n+1}(a) < 1$ for all integers $n \geq 1.$
The harmonic table is a triangular array: $1$ $\frac 12 \qquad \frac 12$ $\frac 13 \qquad \frac 16 \qquad \frac 13$ $\frac 14 \qquad \frac 1{12} \qquad \frac 1{12} \qquad \frac 14$ Where $a_{n,1} = \frac 1n$ and $a_{n,k+1} = a_{n-1,k} - a_{n,k}$ for $1 \leq k \leq n-1.$ Find the harmonic mean of the $1985^{th}$ row.