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I understand that both expresions represent the same sequence of numbers. It starts at $a_{0}=2$ and oscilate around 1 converging to it from up and down.

I have been playing around with the explicit form and I understand how to pass from $\left \{ (-1/2)^n) \right \}^{+\infty}_{n=1}$ to $a_{n}=a_{n-1}(-1/2)^n, a_{0}=1$ that is the same sequence but shifted 1 unit up.

So, it seems that adding one unit to the sequence makes things more complicated when expresing it in recursive form.

My question is, what is the path of thought that someone can use to go from the explicit form $\left \{ 1+(-1/2)^n) \right \}^{+\infty}_{n=1}$ to reach the recursive form $a_{n+2}=1/2(a_{n+1}+a_{n}) , a_{0} = 2 , a_{1}=1/2$?

I´m looking for a way to understand why adding 1 to the sequence $\left \{ (-1/2)^n) \right \}^{+\infty}_{n=1}$ generates the need to express the recursive form of the nth term in function of n-1 and n-2, in order to be able to recognize that need when analyzing similar sequences.

Walquer X. Valles
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    You are dealing with a homogenous recurrence, which can be solved using characteristic polynomials, have a look at this answer as an example ... apply the same technique, but be careful, you have difference initial conditions. – rtybase Jul 25 '18 at 07:42

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In general, a recursion of the form $a_n = k a_{n-1}$ has solution given by $a_n = A k^n$ while a recursion of the form $a_n = k_1 a_{n-1} + k_2 a_{n-2}$ has solution given by $a_n = A c_1^n + B c_2^n$ (or, under certain circumstances, $(n+B) c_1^n$). So, since the sequence you're looking for has two exponential terms ($c_1 = 1$), a second order recurrence might be useful. For more information, you might like to read about linear recurrence relations.

For another perspective, consider $a_n = 1 + \left(-\frac{1}{2}\right)^n$, and our aim is to remove the dependence on $n$ on the right hand side with the help of $a_{n-1}$ and possibly $a_{n-2}$. So, we try considering $a_n - 1 = \left(-\frac{1}{2}\right)^n$ and noticing that $a_{n-1} - 1 = \left(-\frac{1}{2}\right)^{n-1}$ so $-\frac{1}{2}\left(a_{n-1} - 1\right) = \left(-\frac{1}{2}\right)^n$. Matching the right hand sides, we get $2(a_n - 1) = 1 - a_{n-1}$ so $2a_n - 2 = 1 - a_{n-1}$ and the recurrence $2a_n +a_{n-1}= 3$ i.e. $a_n = \frac{3 - a_{n-1}}{2}$, so the extra $a_{n-2}$ wasn't required.

But, this isn't a linear recurrence any more, and perhaps we'd like to make it linear. This is easily done by noticing $2a_{n-1} +a_{n-2}= 3$ also, so

$$2a_n + a_{n-1} = 2a_{n-1} + a_{n-2} \implies a_n = \frac{1}{2}\left(a_{n-1} + a_{n-2}\right).$$

B. Mehta
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