I assume you're asked to solve for
, or find an explicit formula for the n-th term in the sequence.
The sequence is recursively defined by
![\begin{cases} a_1 = 10 \\ a_n = a_(n-1) + 4 & \text{for } n \ge 1 \end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/6qxjfgvt67rvqt7mqizlozpy1oygdsvm3s.png)
By this definition,
![a_(n-1) = a_(n-2) + 4](https://img.qammunity.org/2023/formulas/mathematics/college/acab6413at0uusqyplrrn86nct66gtxvy5.png)
so that by substitution,
![a_n = (a_(n-2) + 4) + 4 = a_(n-2) + 2*4](https://img.qammunity.org/2023/formulas/mathematics/college/3b1t3ytjtx1kjdz5q0r61cthw1d9u2f6l1.png)
and we can repeat this process to find
![a_n = a_(n-3) + 3*4](https://img.qammunity.org/2023/formulas/mathematics/college/ldrscoyo8hsw3juluogjuyd0lml4ae1iwm.png)
![a_n = a_(n-4) + 4*4](https://img.qammunity.org/2023/formulas/mathematics/college/psg3aubvykfxtdm6gelxgxgkqrbk6jsxwj.png)
and so on, down to
![a_n = a_1 + (n-1)*4](https://img.qammunity.org/2023/formulas/mathematics/college/cqb7o670n4yvmcefw0x3qnlifz0d3zjyxx.png)
Then given the first term
, we have
![a_n = 10 + 4(n-1) \implies \boxed{a_n = 4n + 6}](https://img.qammunity.org/2023/formulas/mathematics/college/6ki4oadu8cojipjfdhli8q6fct9yuyz0w7.png)