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Equation is: ${(1/2)(a_n + a_{(n-1)})} = 2n + 5$ when $a_0 = 3$.

So, I got ${a_n = A(-1)^n}$ for the homogenous solution
I got 2n + 1 for a particular solution to ${a_n = -a_{(n-1)} + 4n}$
I got 5 for a particular solution to ${a_n = -a_{(n-1)} + 10}$.

Combined, I used $a_0 = 3$ to get A = -8. So, I ended up with ${a_n = -8(-1)^n + 2n + 11}$

Is that correct? Or, if not, where did I go wrong?

Update: In my original work, I mistakenly added $10$ instead of $5$. So now, I instead got A = -3 with ${a_n = -3(-1)^n + 2n + 6}$. Is this correct?

Jean Marie
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  • $A(0)=3$ should be a $a_0=3$. – Jean Marie Nov 14 '20 at 16:42
  • Combining $2n+1$ and $5$ should yield $2n+6$. Then the corresponding $A$ needs to be changed. But your method is correct. – player3236 Nov 14 '20 at 16:43
  • Gotcha. I added 10 instead of 5. I just updated my work. Does that look correct now? –  Nov 14 '20 at 16:47
  • Yes. To confirm that, we have $(1/2)(a_n + a_{n-1}) = (1/2)(-3(-1)^n + 2n + 6 + -3(-1)^{n+1} + 2n + 4) = 2n+5$, and $a_0 = -3 + 0 + 6 = 3$, which matches the recurrence relation. – player3236 Nov 14 '20 at 16:48

1 Answers1

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Here is a more direct solution. Rewrite your initial relationship under the form:

$$\frac12\left(\underbrace{(a_n-(2n-6))}_{b_n}+\underbrace{(a_{n-1}-(2(n-1)-6))}_{b_{n-1}}\right)=0$$

In this way, you eliminate at once the "non-linearity", giving, as you have done:

$$b_n=A(-1)^n=a_n-(2n-6) \ \iff \ a_n=(2n-6)+A(-1)^n$$

It remains to adjust $A$ in order to comply with the initial value of $a_0$.

We find back your final solution!

Jean Marie
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