Limit of recursive sequence (help for getting the nth term expression) I am trying to find the limit of this sequence (I was able to verify that it is a Cauchy sequence), but I am having problems finding the expression for the nth term. 
$a_{n+2} = \frac{a_n+3a_{n+1}}{4} $
 with  $a_0=0,    a_1=1$
Thanks in advance

For all integer $n\geqslant 0$, we have
$$
  a_{n+2} - a_{n+1} = -\frac{1}{4}(a_{n+1} - a_n).
$$
The sequence $(b_n)_{n\geqslant 0}$ defined by $b_n = a_{n+1} - a_n$ is
geometric with common ratio $-1/4$ and first term $b_0 = 1$, so
$b_n = (-1/4)^n$. Now,
$$
\begin{align*}
  a_n = a_n - a_0 
      = \sum_{i=0}^{n-1} (a_{i+1} - a_i) 
      = \sum_{i=0}^{n-1} \left(\frac{-1}{4}\right)^i 
      = \frac{4}{5}\biggl(1 - \Bigl(\frac{-1}{4}\Bigr)^n\biggr)
\end{align*}
$$
for all $n\geqslant 0$.
