This is the recurrence relation: $a_n=2a_{n-1}+2n, a_2=2$.The way we have found the explicit formulas in my course in the past was to notice some pattern either by finding several values or using the next terms to define $a_n$.However, I'm completely lost on the equation above.
Finding the explicit formula of a given recurrence relation
recurrence-relationssequences-and-series
Related Solutions
Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence as: $$ a_{n + 2} = - 2 a_{n + 1} + 15 a_n $$ Running the recurrence "backwards" gives $a_0 = 6$ (starting at index 0 is nicer all around). Multiply the recurrence by $z^n$, sum over $n \ge 0$, and recognize: \begin{align} \sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\ \sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2} \end{align} to get: $$ \frac{A(z) - 6 - 10 z}{z^2} = - 2 \frac{A(z) - 6}{z} + 15 A(z) $$ Solving for $A(z)$, and writing as partial fractions, gives: $$ A(z) = \frac{6 + 22 z}{1 + 2 z - 15 z^2} = \frac{1}{1 + 5 z} + \frac{5}{1 - 3 z} $$ Two geometric series: $$ a_n = (-5)^n + 5 \cdot 3^n $$ This gives $a_5 = -1910$.
I made a spreadsheet, calculating $a_n$ further than you did, and saw a pattern,
where $a_n$ became close to powers of $2$.
I then made an additional column with the difference between $a_n$ and $2^{n+1}$
and saw a further obvious pattern there.
That led me to hypothesize that $a_n=2^{n+1}-n$, which I then easily proved by induction.
Best Answer
The following manipulations yield a homogeneous linear recurrence relation with constant coefficients:
$$a_n=2a_{n-1}+2n$$
$$a_{n-1}=2a_{n-2}+2(n-1)$$
$$a_n-a_{n-1}=2a_{n-1}-2a_{n-2}+2$$
$$a_n=3a_{n-1}-2a_{n-2}+2$$
$$a_{n-1}=3a_{n-2}-2a_{n-3}+2$$
$$a_n-a_{n-1}=3a_{n-1}-5a_{n-2}+2a_{n-3}$$
$$a_n=4a_{n-1}+5a_{n-2}-2a_{n-3}$$
Can you take it from here?