How are these two signals equal?
What I have tried:
$ x[n] = (-3)(-2)(-2)^{n-1}u[n] + (-2)^{n-1}u[n] $
I am not sure how this leads to the answer.
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1 Answer
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They are indeed equal. The first term is
$$-3(-2)^nu[n]=(-3)(-2)(-2)^{n-1}u[n]=6(-2)^{n-1}u[n]\tag{1}$$
The value of $(1)$ at $n=0$ is $6/(-2)=-3$, so we can rewrite it as
$$-3\delta[n]+6(-2)^{n-1}u[n-1]\tag{2}$$
Adding the second term $(-2)^{n-1}u[n-1]$ results in
$$x[n]=-3\delta[n]+7(-2)^{n-1}u[n-1]\tag{3}$$
