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I have $a_n-a_{n-1}=n$ with some initial conditions $a_0=1, a_2=2, a_3=4, a_4=7$ since there exist $n$, it's inhomogeous equation isn't it?

I know from ODE i can bring the homogenous solution and differentiate it and do a substitution into the original equation to find the particular solution.

But how about this difference equation?

J. W. Tanner
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user516076
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1 Answers1

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Here is a solution:

$a_n-a_0=a_n-a_{n-1}+a_{n-1}-a_{n-2}+a_{n-2}-a_{n-3}+...+a_1-a_0$

$=n+(n-1)+(n-2)+...+(n-(n-1))$

$=n\times n - \frac {(n-1)n}2=\frac{n^2}2+\frac n2,$

so $a_n=\frac{n^2}2+\frac n2+1$.

J. W. Tanner
  • 60,406
  • Thanks it helps me a lot. But how is actually to find the particular solution for this? Somehow, i saw on y*utube about the formula of the general Homogenous Difference Equations like this $$a_h=a_0+\sum_{i=0}^{\infty}fn(i)$$. And it gives me the same answer as you written – user516076 Dec 11 '19 at 12:18