Problems

Prove that for any natural number $a_1>1$ there exists an increasing sequence of natural numbers $a_1,a_2,a_3,\dots$, for which $a_1^2+a_2^2+\cdots +a_k^2$ is divisible by $a_1+a_2+\cdots +a_k$ for all $k\ge 1$.

Is there a sequence of natural numbers in which every natural number occurs exactly once, and for any $k=1,2,3,\dots$ the sum of the first $k$ terms of the sequence is divisible by $k$?

The sequence of numbers $a_1,a_2,\dots$ is given by the conditions $a_1=1$, $a_2=143$ and

for all $n\ge 2$.

Prove that all members of the sequence are integers.

Which term in the expansion $(1+\sqrt{3}{)}^{100}$ will be the largest by the Newton binomial formula?

Prove that in any infinite decimal fraction you can rearrange the numbers so that the resulting fraction becomes a rational number.

$N$ points are given, no three of which lie on one line. Each two of these points are connected by a segment, and each segment is coloured in one of the $k$ colours. Prove that if $N>\lfloor k!e\rfloor$, then among these points one can choose three such that all sides of the triangle formed by them will be colored in one colour.

The Babylonian algorithm for deducing $\sqrt{2}$. The sequence of numbers $\{x_n\}$ is given by the following conditions: $x_1=1$, $x_{n+1}=\frac{1}{2}(x_n+2/x_n)$ ($n\ge 1$).

Prove that $\underset{n\to \mathrm{\infty }}{lim}{x}_n=\sqrt{2}$.

What will the sequence from the previous problem 61297 be converging towards if we choose $x_1$ as equal to $-1$ as the initial condition?

The iterative formula of Heron. Prove that the sequence of numbers $\{x_n\}$ given by the conditions $x_1=1$, $x_{n+1}=\frac{1}{2}(x_n+k/x_n)$, converges. Find the limit of this sequence.

Method of iterations. In order to approximately solve an equation, it is allowed to write $f(x)=x$, by using the iteration method. First, some number $x_0$ is chosen, and then the sequence $\{x_n\}$ is constructed according to the rule $x_{n+1}=f(x_n)$ ($n\ge 0$). Prove that if this sequence has the limit $x\ast =\underset{n\to \mathrm{\infty }}{lim}{x}_n$, and the function $f(x)$ is continuous, then this limit is the root of the original equation: $f(x^{\ast })=x^{\ast }$.