What is the eighth term given the sequence: +896,−448,+224,−112,⋯ ?

Step 1
Let any position count be i
Let any term be ${\displaystyle a_i}$
So
${\displaystyle a_1\to }$ first term
${\displaystyle a_2\to }$ second term
${\displaystyle a_i\to }$ its term
Two things I notice:
Step 1
Alternates between positive and negative starting at positive.
So this can be achieved by
${\displaystyle a_i\to a_1\times {(-1)}^2}$
${\displaystyle a_i\to a_2\times {(-1)}^3}$
${\displaystyle a_i\to a_3\times {(-1)}^4}$
${\displaystyle \Rightarrow a_i\to a_i\times {(-1)}^{i+1}}$
Step 2
Each term is half the previous
So we have
${\displaystyle {(-1)}^2\times a_1}$
${\displaystyle a_2={(-1)}^3\times \frac{1}{2}\times a_1}$
${\displaystyle a_3={(-1)}^4\times \frac{1}{2}\times \frac{1}{2}\times a_1}$
This implies ${\displaystyle a_1={(-1)}^{i+1}\times {\left(\frac{1}{2}\right)}^{i-1}\times a_1}$
Step 3
${\displaystyle a_i=896{(-1)}^{i+1}{\left(\frac{1}{2}\right)}^{i-1}}$
Thus
${\displaystyle a_8=896{(-1)}^{8+1}{\left(\frac{1}{2}\right)}^{8-1}=-7}$