Series Question

Let the series  $a_{n+1} =1+\sum{a_n}+a_n$ if $a_2=7$ what is $a_5$ ?

$\begin{array}{rcll} a_{n+1} &=& 1+\sum{a_n}+a_n \\\\ a_2 &=& 1 + a_1 + a_1\\ \mathbf{a_2} &\mathbf{=}& \mathbf{1 + 2a_1} \qquad & | \qquad a_2 = 7\\ 7 &=& 1 + 2a_1 \\ 6 &=& 2a_1 \\ 3 &=& a_1 \\ \mathbf{a_1} &\mathbf{=}& \mathbf{3}\\ \end{array}$

$\begin{array}{rcll} a_3 &=& 1 + a_1 + a_2 + a_2\\ a_3 &=& 1 + a_1 + 2a_2 \qquad & | \quad \small{a_2 = 1+2a_1}\\ a_3 &=& 1 + a_1 + 2\cdot(1+2a_1)\\ a_3 &=& 1 + a_1 + 2 + 4a_1\\ \mathbf{a_3} &\mathbf{=}& \mathbf{3 + 5a_1}\qquad & | \qquad a_1 = 3\\ a_3 &=& 3 + 5\cdot 3\\ \mathbf{a_3} &\mathbf{=}& \mathbf{18}\\ \end{array}$

$\begin{array}{rcll} a_4 &=& 1 + a_1 + a_2 + a_3 + a_3\\ a_4 &=& 1 + a_1 + a_2 +2a_3 \qquad & | \quad \small{a_2 = 1+2a_1 \quad a_3 = 3 + 5a_1}\\ a_4 &=& 1 + a_1 + 1+2a_1+ 2\cdot( 3 + 5a_1)\\ a_4 &=& 1 + a_1 + 1+2a_1+ 6 + 10a_1\\ \mathbf{a_4} &\mathbf{=}& \mathbf{8 + 13a_1}\qquad & | \qquad a_1 = 3\\ a_4 &=& 8 + 13\cdot 3\\ \mathbf{a_4} &\mathbf{=}& \mathbf{47}\\ \end{array}$

$\begin{array}{rcll} a_5 &=& 1 + a_1 + a_2 + a_3 + a_4 + a_4\\ a_5 &=& 1 + a_1 + a_2 + a_3 + 2a_4 \qquad & | \quad \small{a_2 = 1+2a_1 \quad a_3 = 3 + 5a_1 \quad a_4 = 8 + 13a_1}\\ a_5 &=& 1 + a_1 + 1+2a_1+ 3 + 5a_1 +2\cdot (8 + 13a_1)\\ a_5 &=& 1 + a_1 + 1+2a_1+ 3 + 5a_1 + 16 + 26a_1\\ \mathbf{a_5} &\mathbf{=}& \mathbf{21 + 34a_1}\qquad & | \qquad a_1 = 3\\ a_5 &=& 21 + 34\cdot 3\\ \mathbf{a_5} &\mathbf{=}& \mathbf{123}\\ \end{array}$

Nice thanks!