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Some linear recurrence relations that do not have constant coefficients can be systematically solved. This is the case for recurrence relations of the form $f (n)a_{n} = g(n)a_{n-1} + h(n).$ Exercises $48–50$ illustrate this.

  1. Show that the recurrence relation $f (n)a_{n} = g(n)a_{n-1} + h(n),$ for $n \geq 1,$ and with $a_{0} = C,$ can be reduced to a recurrence relation of the form $b_{n} = b_{n-1} + Q(n)h(n),$ where $b_{n} = g(n + 1)Q(n + 1)a_{n},$ with $Q(n) = \dfrac{(f (1)f (2) \dots f (n - 1))}{(g(1)g(2) \dots g(n))}.$
  2. Use part $(A)$ to solve the original recurrence relation to obtain $a_{n} = \dfrac{C +\displaystyle{} \sum_{i = 1}^{n}Q(i)h(i)}{g(n + 1)Q(n + 1)}$
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