A sequence $(a_n)$ of real numbers is defined by $a_0=1$, $a_1=2015$ and for all $n\geq1$, we have $$a_{n+1}=\frac{n-1}{n+1}a_n-\frac{n-2}{n^2+n}a_{n-1}.$$ Calculate the value of $\frac{a_1}{a_2}-\frac{a_2}{a_3}+\frac{a_3}{a_4}-\frac{a_4}{a_5}+\ldots+\frac{a_{2013}}{a_{2014}}-\frac{a_{2014}}{a_{2015}}$.
Problem
Source: 2015 CentroAmerican Math Olympiad #2
Tags: OMCC, algebra
A_Gappus
28.06.2015 04:00
Easy to prove $\frac{a_n}{a_{n+1}}=n+1$ by induction
Indeed, it's correct for $n=1,2,...$. Assume $ \frac{a_{n-1}}{a_n}=n$. Combine to condition $ a_{n+1}=\frac{n-1}{n+1}a_{n}-\frac{n-2}{n^2+n}a_{n-1}\Rightarrow (n+1)a_{n+1}=a_n\Rightarrow \frac{a_n}{a_{n+1}}=n+1$
So, $(\frac{a_1}{a_2}-\frac{a_2}{a_3})+....+(\frac{a_{2013}}{a_{2014}}-\frac{a_{2014}}{a_{2015}})=1\times 1007=1007$
RMACR7LP
28.06.2015 10:39
Remember that a_1 is 2015, so the first parenthesis is not equal to one
marioalejandroruiz
28.02.2016 05:24
The real answer is 3021 because of what "RMACR7LP" said.
congthanh_lekhiet
25.11.2018 17:45
It's easy to prove that: $a_{n+1}=\frac{1}{n+1}.a_{n}$ ($n\geqslant 2$) by induction. On the other hand, we have: $a_{2}=\frac{1-1}{1+1}a_{1}-\frac{1-2}{1^{2}+1}a_{0}=\frac{1}{2}$ Hence, $\frac{a_{1}}{a_{2}}-\frac{a_{2}}{a_{3}}+\frac{a_{3}}{a_{4}}-\frac{a_{4}}{a_{5}}+...+\frac{a_{2013}}{a_{2014}}-\frac{a_{2014}}{a_{2015}}$ = $\frac{2015}{\frac{1}{2}}-3+4-5+...+2014-2015=3021$
WolfusA
29.11.2018 15:56
FOr the sake of completeness we should prove $a_i\neq 0$.