A sequence ($a_n$) is given by $a_0 = 0, a_1 = 1$ and $a_{k+2} = a_{k+1} +a_k$ for all integers $k \ge 0$. Prove that the inequality $\sum_{k=0}^n \frac{a_k}{2^k}< 2$ holds for all positive integers $n$.

Let $x_1,x_2,...,x_n$ be real numbers in an interval $[m,M]$ such that $\sum_{i=1}^n x_i = 0$. Show that $\sum_{i=1}^n x_i ^2 \le -nmM$

Assume the $m$ is a given integer greater than $ 1$. Find the largest number $C$ such that for all $n \in N$ we have $$\sum_{1\le k \le m ,\,\, (k,m)=1}\frac{1}{k}\ge C \sum_{k=1}^{m}\frac{1}{k}$$

For any nonempty set $S$, we define $\sigma(S)$ and $\pi(S)$ as sum and product of all elements from set $S$, respectively. Prove that $a)$ $\sum \limits_{} \frac{1}{\pi(S)} =n$ $b)$ $\sum \limits_{} \frac{\sigma(S)}{\pi(S)} =(n^2+2n)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)(n+1)$ where $\sum$ denotes sum by all nonempty subsets $S$ of set $\{1,2,...,n\}$

Let there be a $n\times n$ board. Write down $0$ or $1$ in all $n^2$ squares. For $1 \le k \le n$, let $A_k$ be the product of all numbers in the $k$th row. How many ways are there to write down the numbers so that $A_1 + A_2 + ... + A_n$ is even?

We write $\{a,b,c\}$ for the set of three different positive integers $a, b$, and $c$. By choosing some or all of the numbers a, b and c, we can form seven nonempty subsets of $\{a,b,c\}$. We can then calculate the sum of the elements of each subset. For example, for the set $\{4,7,42\}$ we will find sums of $4, 7, 42,11, 46, 49$, and $53$ for its seven subsets. Since $7, 11$, and $53$ are prime, the set $\{4,7,42\}$ has exactly three subsets whose sums are prime. (Recall that prime numbers are numbers with exactly two different factors, $1$ and themselves. In particular, the number $1$ is not prime.) What is the largest possible number of subsets with prime sums that a set of three different positive integers can have? Give an example of a set $\{a,b,c\}$ that has that number of subsets with prime sums, and explain why no other three-element set could have more.

For every four-digit number, we take the product of its four digits. Then we add all of these products together . What is the result? ( G Galperin)

Consider ten concentric circles and ten rays as in the following figure. At the points where the inner circle is intersected by the rays write successively, in direction clockwise, the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. In the next circle we write the numbers $11, 12, 13, 14, 15, 16, 17, 18, 19,20$ successively, and so on successively until the last round were we write the numbers $91, 92, 93, 94, 95, 96, 97, 98, 99, 100$ successively. In this orde, the numbers $1, 11, 21, 31, 41, 51, 61, 71, 81, 91$ are in the same ray, and similarly for the other rays. In front of $50$ of those $100$ numbers, we use the sign ''$-$'' such as: a) in each of the ten rays, exist exactly $5$ signs ''$-$'' , and also b) in each of the ten concentric circles, to be exactly $5$ signs ''$-$''. Prove that the sum of the $100$ signed numbers that occur, equals zero. [img]https://cdn.artofproblemsolving.com/attachments/9/d/ffee6518fcd1b996c31cf06d0ce484a821b4ae.gif[/img]

Find the number of all real functions $f$ which map the sum of $n$ elements into the sum of their images, such that $f^{n-1}$ is a constant function and $f^{n-2}$ is not. Here $f^0(x) = x$ and $f^k = f \circ f^{k-1}$ for $k \ge 1$.

Calculate $$\frac{(2^1+3^1)(2^2+3^2)(2^4+3^4)(2^8+3^8)...(2^{2048}+3^{2048})+2^{4096}}{3^{4096}}$$

Let $n$ be a given positive integer. Find the solution set of the equation $\sum_{k=1}^{2n} \sqrt{x^2 -2kx + k^2} =|2nx - n - 2n^2|$  

Suppose that $a_1,a_2,...,a_n$ are integers such that $a_1 +2^ia_2 +3^ia_3 +...+n^ia_n = 0$ for $i = 1,2,...,k -1$, where $k \ge 2$ is a given integer. Prove that $a_1+2^ka_2+3^ka_3+...+n^ka_n$ is divisible by $k!$.

Compute the sum $$\sum_{k=0}^{n}\frac{(2n)!}{k!^2(n-k)!^2}.$$

There are $2n$ distinct points on a circle. The numbers $1$ through $2n$ are randomly assigned to this one points distributed. Each point is connected to exactly one other point, so that no of the resulting connecting routes intersect. If a segment connects the numbers $a$ and $b$, so we assign the value $ |a - b|$ to the segment . Show that we can choose the routes such that the sum of these values ​​results $n^2$.

Is it true that every positive integer greater than $100$ is a sum of $4$ positive integers such that each two of them have a common divisor greater than $1$?

Does there exist a four-digit positive integer with different non-zero digits, which has the following property: if we add the same number written in the reverse order, then we get a number divisible by $101$?