Let [tex]a_n[/tex] denote the [tex]n[/tex]-th term in the sequence. By checking the forward differences, we observe
[tex]a_2 - a_1 = 5 - 1 = 4 = 4\cdot3^0[/tex]
[tex]a_3 - a_2 = 17 - 5 = 12 = 4\cdot3^1[/tex]
[tex]a_4 - a_3 = 53 - 17 = 36 = 4\cdot3^2[/tex]
[tex]a_5 - a_4 = 161 - 53 = 108 = 4\cdot3^3[/tex]
The pattern is
[tex]\begin{cases} a_1 = 1 \\ a_n = a_{n-1} + 4\cdot3^{n-2} & \text{for } n\ge1 \end{cases}[/tex]
So we find
[tex]a_6 = a_5 + 4\cdot3^{6-2} = 161 + 4\cdot3^4 = \boxed{485}[/tex]
[tex]a_7 = a_6 + 4\cdot3^{7-2} = 485 + 4\cdot3^5 = \boxed{1457}[/tex]
We also could have solved for [tex]a_n[/tex] first. By substitution,
[tex]a_{n-1} = a_{n-2} + 4\cdot3^{n-3} \\\\ ~~~~ \implies a_n = a_{n-2} + 4 \left(3^{n-2} + 3^{n-3}\right)[/tex]
[tex]a_{n-2} = a_{n-3} + 4\cdot3^{n-4} \\\\ ~~~~ \implies a_n = a_{n-3} + 4 \left(3^{n-2} + 3^{n-3} + 3^{n-4}\right)[/tex]
[tex]a_{n-3} = a_{n-4} + 4\cdot3^{n-5} \\\\ ~~~~ \implies a_n = a_{n-4} + 4 \left(3^{n-2} + 3^{n-3} + 3^{n-4} + 3^{n-5}\right)[/tex]
and so on. After so many iterations of this, we see the pattern
[tex]a_n = a_{n-k} + 4 \displaystyle \sum_{\ell=1}^k 3^{n-2-(\ell-1)} = a_{n-k} + 4 \cdot 3^{n-1} \sum_{\ell=1}^k \frac1{3^\ell}[/tex]
so that for [tex]k=n-1[/tex], we get
[tex]\displaystyle a_n = a_1 + 4 \cdot 3^{n-1} \sum_{\ell=1}^{n-1} \frac1{3^\ell}[/tex]
Let [tex]S_{n-1}[/tex] be the remaining sum. We have
[tex]S_{n-1} = \dfrac13 + \dfrac1{3^2} + \dfrac1{3^3} + \cdots + \dfrac1{3^{n-1}}[/tex]
[tex]\dfrac13 S_{n-1} = \dfrac1{3^2} + \dfrac1{3^3} + \dfrac1{3^4} + \cdots + \dfrac1{3^n}[/tex]
[tex]\implies \dfrac23 S_{n-1} = \dfrac13 - \dfrac1{3^n}[/tex]
[tex]\implies S_{n-1} = \dfrac12 \left(1 - \dfrac1{3^{n-1}}\right)[/tex]
and so
[tex]\displaystyle a_n = 1 + 4 \cdot 3^{n-1} \cdot \dfrac12 \left(1 - \dfrac1{3^{n-1}}\right) \\\\ a_n = 2\cdot3^{n-1} - 1[/tex]
Then
[tex]a_6 = 2\cdot3^{6-1}-1 = 485[/tex]
[tex]a_7 = 2\cdot3^{7-1}-1 = 1457[/tex]
