A number theory problem by Guru Prasaadh

I did it the long way, hoping someone will provide a simpler solution.

I checked the differences of successive terms ( $\space \Delta _n = a_n - a_{n-1}$ ), the second differences ( $\space \Delta _n ^2 = \Delta_n - \Delta_{n-1}$ ) and the third differences ( $\space \Delta _n ^3 = \Delta_n^2 - \Delta_{n-1}^2$ ). And the results are as follows:

$\begin{matrix} n & a_n & \Delta_n & \Delta_n^2 & \Delta_n^3 \\ 1 & 1 & & & \\ 2 & 0 & -1 & & \\ 3 & 3 & 3 & 4 & \\ 4 & 16 & 13 & 10 & 6 \\ 5 & 45 & 29 & 16 & 6 \\ 6 & 96 & 51 & 22 & 6 \end{matrix}$

It is noted that for $n \ge 4$ , $\Delta_n^3 = 6$ , a constant.

We know that:

$\Delta _n = a_n - a_{n-1}$

$\Delta _n ^2 = \Delta_n - \Delta_{n-1} = (a_n - a_{n-1}) - (a_{n-1} - a_{n-2}) = a_n - 2a_{n-1} + a_{n-2}$

$\Delta _n ^3 = \Delta_n^2 - \Delta_{n-1}^2 = (a_n - 2a_{n-1} + a_{n-2}) - (a_{n-1} - 2a_{n-2} + a_{n-3})$ $\quad \quad \quad \quad \quad \quad \quad \quad = a_n - 3a_{n-1} + 3a_{n-2} - a_{n-3}$

Note that in general: $\space \Delta _n ^k = \displaystyle \sum _{i=0} ^k {\begin{pmatrix} k \\ i \end{pmatrix} a_{n-i}}$

Therefore,

$\Delta_n^3 = a_n - 3a_{n-1} + 3a_{n-2} - a_{n-3} = 6$

$\Rightarrow a_n = 6 + 3a_{n-1} - 3a_{n-2} + a_{n-3}$

$\Rightarrow a_7 = 6 + 3(96) - 3(45) + 16 = 6 + 288 -135 + 16 = \boxed{175}$

The general term can be written as - ((-1+r)^2)*(r+1) with r varying from 0 to whichever value is wanted.