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I have a problem with this type of non-homogeneous equation.

Find the solution of recurrence equation:

$2 A_{n+1} = 3A_{n}-n+2$

$A_{0} = 1$

I know the idea behind the problem when the particular part is in the form that n is an exponent of some constant, but i don't know how to act with this.

Mhenni Benghorbal
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Filip
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1 Answers1

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$$ \frac{A_{n+1}}{\left(\frac{3}{2}\right)^{n+1}} = \frac{A_{n}}{\left(\frac{3}{2}\right)^{n}} + (1 - \frac{n}{2}) \left(\frac{2}{3}\right)^{n+1} $$

So you can add this equation from 1 to n.

Generally speaking, for recurrence equation $ a_{n+1} = p a_{n} + f(n) $, we let $ b_{n} = \frac{a_{n}}{p^n} $, so that $ b_{n+1} = b_{n} + \frac{f(n)}{p^{n+1}} $.

$$ \begin{aligned} b_{n} - b_{n-1} & = \frac{f(n-1)}{p^{n}} \\ b_{n-1} - b_{n-2} & = \frac{f(n-2)}{p^{n-1}} \\ &... \\ b_{k} - b_{k-1} & = \frac{f(k-1)}{p^{k}}\\ & ... \\ b_{2} - b_{1} & = \frac{f(1)}{p^{2}} \end{aligned} $$

So we get $ b_{n} = b_{1} + \sum_{k=1}^{n-1} \frac{f(k)}{p^{k+1}} $

easymath3
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