Sequence which changes its sign like this Which is the general formula of this sequence?
$$ x_0 = -1$$
$$ x_{n+1} = ((-1)^n*X_n)/2^n$$
What baffles me more is the sign which is like this: $--++--++--++--++\cdots$
I've been wondering how the sign could change like that, any ideas?
 A: Let $\,a_n=(-1)^{\lfloor n/2 \rfloor+1}\,$, then by direct computation $a_0=a_1=-1\,$, $a_2=a_3=1\,$, and:
$$
a_{n+4}=(-1)^{\lfloor (n+4)/2 \rfloor+1} = (-1)^{\lfloor n/2 + 2 \rfloor+1}=(-1)^2 \cdot (-1)^{\lfloor n/2 \rfloor+1}=(-1)^{\lfloor n/2 \rfloor+1} = a_n
$$
Therefore the sequence is periodic with period $\,4\,$, and so the first four values $\,-1,-1,1,1\,$ keep repeating indefinitely.
A: You can easly prove by induction that $\forall n\in\mathbb{N},x_n\neq0$.
Now let $n\ge 1$. We have $\forall k\in\{0,..,n-1\},x_{k+1}=\left(\dfrac{-1}{2}\right)^kx_k$. Thus
\begin{aligned}
\prod_{k=0}^{n-1}x_{k+1}&=\prod_{k=0}^{n-1}\left(\dfrac{-1}{2}\right)^k x_k\\
\prod_{k=1}^nx_k&=\left(\dfrac{-1}{2}\right)^{\left(\sum_{k=0}^{n-1}k\right)}\prod_{k=0}^{n-1}x_k.
\end{aligned}
You can eliminate $\prod_{k=1}^{n-1}x_k$ and we have$\sum_{k=0}^{n-1}k=\dfrac{(n-1)n}{2}$. Therefore
\begin{aligned}
x_{n+1}&=\left(\dfrac{-1}{2}\right)^{\frac{(n-1)n}{2}}x_0\\
&=-\left(\dfrac{-1}{2}\right)^{\frac{(n-1)n}{2}}.
\end{aligned}
Hence a general formula that gives your described sign is $(-1)^{\frac{(n-1)n}{2}}$. Note that even if this goes beyond your math level, you can prove the formula we got for $x_n$ by induction.
A: Let $n=2k $ be even.  Then $(-1)^n=1$ and $x_{n+1}$ will have the same sign as $x_n$.
Let $n=2k+1$ be odd.  Then $(-1)^n=-1$ and $x_{n+1} $ will be the opposite sign as $x_n $.
Therefore $x_{n+2}$ will always be the opposite sign as $x_{n} $.  Because if $n $ is even then $x_{n+1}$ will have the same sign, and $n+1$ is odd  so $x_{n+1}$ has the opposite sign.  If $n$ is odd then $x_{n+1}$ has the opposite sign, and $n+1$ is even so $x_{n+1} $ also has the opposite sign.  
And $x_{n+4} $ will have the same sign as $x_{n+4} $ will have the opposite of $x_{n+2} $ which is the opposite of $x_n$.
So $x_0 = -$ by definition.
$x_1=-$ because 0 is even so sign stays the same.
$x_2=+$ because $1$ is odd so sign changes.
$x_3=+$ because $2$ is even so sign stays the same
$x_4=-$ and pattern repeats forever as pattern repeats four every fourth term.
