Re: Dhaka-2 Higher Secondary 2013 / 8
Forum rules
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
A sequence of real numbers $x_i$ is defined such that its first term $x_0 = 1$ and for $n\geq 1, x_n = \sqrt {2x_{n-1}+4}$. The terms of this sequence are never larger than a certain real number. This real number can be written as $a+b\sqrt{c}$. Find $a+b+c$.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
-
asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: Dhaka-2 Higher Secondary 2013 / 8
I think it should be $\sqrt{a+b\sqrt{c}}$.All the terms of this sequence are never larger than $\sqrt{4+2\sqrt{6}}$.Labib wrote:A sequence of real numbers $x_i$ is defined such that its first term $x_0 = 1$ and for $n\geq 1, x_n = \sqrt {2x_{n-1}+4}$. The terms of this sequence are never larger than a certain real number. This real number can be written as $a+b\sqrt{c}$. Find $a+b+c$.
Last edited by asif e elahi on Wed Jan 29, 2014 5:47 pm, edited 1 time in total.
Re: Re: Dhaka-2 Higher Secondary 2013 / 8
I just copied the problem from the problem set.
If you think the statement is wrong, why don't you post your solution?
It will help everyone.
If you think the statement is wrong, why don't you post your solution?
It will help everyone.
Last edited by Labib on Wed Jan 29, 2014 11:27 pm, edited 1 time in total.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
-
asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
- Location:Sylhet,Bangladesh
Re: Re: Dhaka-2 Higher Secondary 2013 / 8
For $x_{i}>\sqrt{5}+1$
$(x_{i}-1)^{2}>5$
or $2x_{i}+4<x_{i}^{2}$
or $x_{i+1}=\sqrt{2x_{i}+4}<x_{i}$
So the terms of this sequence decrease as i increases.
It is easy to find $x_{2}=\sqrt{2\sqrt{6}+4}>\sqrt{5}+1$
So $x_{0}<x_{1}<x_{2}>x_{3}>x_{4}..............$
So $x_{2}$ is the greatest term in this sequence .This implies the real number is $\sqrt{2\sqrt{6}+4}$
$(x_{i}-1)^{2}>5$
or $2x_{i}+4<x_{i}^{2}$
or $x_{i+1}=\sqrt{2x_{i}+4}<x_{i}$
So the terms of this sequence decrease as i increases.
It is easy to find $x_{2}=\sqrt{2\sqrt{6}+4}>\sqrt{5}+1$
So $x_{0}<x_{1}<x_{2}>x_{3}>x_{4}..............$
So $x_{2}$ is the greatest term in this sequence .This implies the real number is $\sqrt{2\sqrt{6}+4}$