mathimatikoi.org

Let \( a_n , n \in \mathbb{N} \) be a real sequence such that:

$$a_n=\frac{n+1}{n-1}\sum_{i=1}^{n-1}a_i, \; n \geq 2$$

Evaluate \( a_{2013} \) or even better find \( a_n \) for all \( n \geq 2 \).

For the sequence \(\{{a_n}\}_{n\in\mathbb{N}}\) we have that \[a_n=\frac{n+1}{n-1}\sum_{i=1}^{n-1}a_i, \quad n \geqslant 2\,.\]
We'll prove inductively that \[a_n=(n+1)\,2^{n-2}a_1\,,\quad n \geqslant 2\,.\] \(\color{gray}\bullet\) For \(n=2\) it is trivial: \(a_2=\frac{2+1}{2-1}\,a_1=3a_1=(2+1)\,2^{2-2}a_1\,.\)

\(\color{gray}\bullet\) We assume that it holds for \(n=k\), i.e. \[a_k=\frac{k+1}{k-1}\sum_{i=1}^{k-1}a_i=(k+1)\,2^{k-2}a_1\quad (1)\,.\] We'll prove that also holds for \(n=k+1\), i.e. \[a_{k+1}=\frac{k+2}{k}\sum_{i=1}^{k}a_i=(k+2)\,2^{k-1}a_1\,.\]
\begin{align*}
a_{k+1}&=\frac{k+2}{k}\sum_{i=1}^{k}a_i\\
&=\frac{k+2}{k}\biggl({\sum_{i=1}^{k-1}a_i+a_k}\biggr)\\
&\stackrel{(1)}{=}\frac{k+2}{k}\bigl({(k-1)\,2^{k-2}a_1+(k+1)\,2^{k-2}a_1}\bigr)\\
&=\frac{k+2}{k}\,2k\,2^{k-2}a_1\\
&=(k+2)\,2^{k-1}a_1\,.\quad Q.E.D.
\end{align*}
So \(a_{2013}=2014\cdot2^{2011}\,a_1\,.\)