Answer
$8-16+32-64+........+(-1)^{n+1}2^{n}$
Work Step by Step
We see that there are $n-2$ terms, as the index k changes from $3$ to $n$. The index $k$ indicates how the terms are formed.
Thus, we write the sum for the $n$ term as follows:
$\displaystyle \sum_{k=3}^{n}(-1)^{k+1}2^{k}=$
$=(-1)^{3+1}2^{3}+(-1)^{4+1}2^{4}+(-1)^{5+1}2^{5}+......+(-1)^{n+1}2^{n} \\=(-1)^{4}2^{3}+(-1)^{5}2^{4}+(-1)^{6}2^{5}+.....+(-1)^{n+1}2^{n}$
We find that we have alternating sign. The value of $(-1)^{n}$ is $-1$ when $n$ is odd and $+1$ when $n$ is even.
$=2^{3}-2^{4}+2^{5}-2^{6}+.......+(-1)^{n+1}2^{n} \\ = 8-16+32-64+........+(-1)^{n+1}2^{n}$
