One way to do this would be to replace $a_k$ by $b_k = a_k + c(-1)^k$ (where $c$ is a constant to be chosen later). Then $a_k = b_k - c(-1)^k$, and the recurrence equation for $a_k$ becomes $$b_k - c(-1)^k = 3(b_{k-1} - c(-1)^{k-1}) - (b_{k-2} - c(-1)^{k-2}) - 2(-1)^k,$$ $$b_k = 3b_{k-1} - b_{k-2} + (-1)^k(c + 3c + c - 2).$$ Now choose $c$ so that $5c-2=0$ (so $c = \frac25$). That eliminates the awkward $(-1)^k$ term from the $b_k$ equation, which you should now be able to solve. Having found the answer for $b_k$, you then have $a_k = b_k - \frac25(-1)^k$.goohu said:
Sequence inhomogeneous recursion is a mathematical concept that involves recursively generating a sequence of values where each value is dependent on the previous values in the sequence and also on an external factor or variable.
Regular recursion involves a function calling itself repeatedly to solve a problem, while sequence inhomogeneous recursion involves generating a sequence of values based on a recursive formula that takes into account both the previous values in the sequence and an external factor.
Sequence inhomogeneous recursion can be used to model and solve complex problems that involve a combination of recursive and non-recursive factors. It is often used in mathematical and scientific fields to describe and analyze real-world phenomena.
One example is the Fibonacci sequence, where each number is the sum of the two previous numbers in the sequence. Another example is the population growth of a species, where the population in each generation depends on the previous generations and environmental factors.
Sequence inhomogeneous recursion can become computationally expensive and time-consuming for large and complex problems. It also requires a clear understanding of the recursive formula and external factors, as well as careful handling of boundary conditions to avoid infinite loops.