We know that \(a_6 = 2a_5 - a_4\text{.}\) So to find \(a_6\) we need to find \(a_5\) and \(a_4\text{.}\) Well
so if we can only find \(a_3\) and \(a_2\) we would be set. Of course
so we only need to find \(a_1\) and \(a_0\text{.}\) But we are given these. Thus
Note that now we can guess a closed formula for the \(n\)th term of the sequence: \(a_n = n+3\text{.}\) To be sure this will always work, we could plug in this formula into the recurrence relation:
That is not quite enough though, since there can be multiple closed formulas that satisfy the same recurrence relation; we must also check that our closed formula agrees on the initial terms of the sequence. Since \(a_0 = 0 + 3 = 3\) and \(a_1 = 1+3 = 4\) are the correct initial conditions, we can now conclude we have the correct closed formula.