How to get from $1 + (-1)^{n+1}$ to $1 + [((−1)^{n }) − 1] (−1)$

Question

I need some help with the algebra here.

I have the following explanation, and I really can't follow the algebra. Could you also maybe give me some tips on how to think about such problems.

$a_n = 1 + (−1)^n$

Basis Step: Specify $a_1$ by $a_1 = 1 + (−1)^1 = 0$.

Recursive Step: Give a rule for finding $a_{n+1}$ from $a_n$, for $n ≥ 1$:

$a_{n+1} = 1 + (−1)^{n+1}$

$= 1 + (−1)^n(−1)$

$= 1 + [((−1)^n + 1) − 1] (−1)$

$= 1 + (a_n − 1)(−1)$

$= 2 − a_n$

Thank you!

Answer 1

To get from $(-1)^n(-1)$ to $[((-1)^n + 1) - 1](-1)$, you need to add $0$ (which doesn't change anything) in the sense that you add and subtract $1$. In particular, $\color{#aa0000}{ 1 - 1} = 0$, so $$(-1)^n(-1) = [(-1)^n + \color{#aa0000}{1 - 1}](-1) = [((-1)^n + 1) - 1](-1)$$

Does that make sense?

Answer 2

We know that

$$ a^xa^y = a^{x+y} $$

So we can rewrite $$ 1+(-1)^{n+1} = 1 + (-1)^n(-1) $$

Now pull out $-1$ from both terms

$$ = -1 \left( -1 + (-1)^n \right) $$ Rearranging the insides we see $$ = (-1) \left( (-1)^n - 1 \right) $$ Which is what you have

Let me know if you want an explanation of the specific induction proof you have given EDIT: I decided to give an explanation :P So $$ 2-a_n = 2 - ( 1 + (-1)^n) = 1 - (-1)^n = -1 ( -1 + (-1)^n ) $$ In the least this is what I believe that proof is trying to get at

Let me know if any other help is needed

Answer 3

The problem is to express the sequence recursively, given a closed-form expression for it (or so I think ... this feels backwards). To be specific, we wish to express $a_{n+1}$ in terms of $a_n$. In other words, we want to find an $a_n$ in the closed-form expression for $a_{n+1}$.

Keep in mind that the $a_n$ that we are looking for is equal to $1 + (-1)^n$

$a_{n+1} = 1 + (-1)^{n+1}$ holds by the closed form given above.

$= 1 + (-1)^n(-1)$ holds by the definition of exponentiation.

$= 1 + [((-1)^n + 1) - 1](-1)$ holds because we're just adding and subtracting $1$. Doing this allows for the substitution of $a_n$ below.

$= 1 + [a_n - 1](-1)$ holds by the given expression for $a_n$.

$= 2 - a_n$ by carrying out the multiplication and rearranging things.

And now we have the sequence on a recursive form (with $a_1 = 0$), which is what we were looking for.