Recurrent sequence

Let $a_n$ be a real sequence such that $a_1=2$ and:
$$a_n=\frac{n+1}{n-1}\sum_{i=1}^{n-1}a_i, \;\; n  \geq 2$$

Express $a_n$ in an inductive form.

Solution

 
Successively we have:

$$\begin{aligned}
a_n &=\frac{n+1}{n-1}\sum_{i=1}^{n-1}a_i \\
 &= \frac{n+1}{n-1}\left [ \sum_{i=1}^{n-2}a_i +a_{n-1} \right ]=\frac{n+1}{n-1}\left [ \sum_{i=1}^{n-1}a_i +\frac{n}{n-2}\sum_{i=1}^{n-2}a_i \right ]\\
 &= \frac{n+1}{n-1}\cdot \frac{2n-2}{n-2}\sum_{i=1}^{n-2}a_i= 2\cdot \frac{n+1}{n-2}\sum_{i=1}^{n-2}a_i\\
 &= 2\cdot \frac{n+1}{n-2}\left [ \sum_{i=1}^{n-3}a_i +a_{n-2} \right ]\\
 &= \cdots\\
 &=2^{n-2}(n+1)a_1 = 2^{n-2}(n+1)\cdot 2 \\
 &=2^{n-1}(n+1)
\end{aligned}$$

hence $a_n=2^{n-1} (n+1), \; n \geq 1$.