MathDB

1

Part of 1993 Romania Team Selection Test

Problems(4)

complex sequence z_n = (1+i)(2+i)...(n+i)

Source: Romania BMO TST 1993 p1

2/17/2020
Consider the sequence zn=(1+i)(2+i)...(n+i)z_n = (1+i)(2+i)...(n+i). Prove that the sequence ImIm znz_n contains infinitely many positive and infinitely many negative numbers.
compexSequencealgebra
Nice!

Source:

8/26/2006
Find max. numbers AA wich is true ineq.: xy2+z2+yx2+z2+zx2+y2≥A\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{x^{2}+z^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}\geq A x,y,zx,y,z are positve reals numberes! :wink:
inequalities unsolvedinequalities
f((x+y)/2) < (f(x)+ f(y))/2 , a_n = f(n) does not contain arithmetic progression

Source: Romania IMO TST 1993 2.1

2/17/2020
Let f:R+→Rf : R^+ \to R be a strictly increasing function such that f(x+y2)<f(x)+f(y)2f\left(\frac{x+y}{2}\right) < \frac{f(x)+ f(y)}{2} for all x,y>0x,y > 0. Prove that the sequence an=f(n)a_n = f(n) (n∈Nn \in N) does not contain an infinite arithmetic progression.
arithmetic sequencefunctionIncreasingalgebra
express sequence x_n as a function of n

Source: Romania IMO TST 1993 3.1

2/17/2020
Define the sequence (xnx_n) as follows: the first term is 11, the next two are 2,42,4, the next three are 5,7,95,7,9, the next four are 10,12,14,1610,12,14,16, and so on. Express xnx_n as a function of nn.
Sequencealgebra