Answer
False.
See counterexample below.
Work Step by Step
Let n be an even number, say 4.
We will build a counterexample using
$(-1)^{odd}=-1,\qquad (-1)^{even}=+1$
Let $a_{i}=b_{i}=(-1)^{i}$
Then, the sum on the LHS is
$\displaystyle \sum_{i=1}^{4}a_{i}b_{i}=\sum_{i=1}^{4}(-1)^{i}(-1)^{i}$
$=\displaystyle \sum_{i=1}^{4}[(-1)(-1)]^{i}$
$=\displaystyle \sum_{i=1}^{4}1^{i}$
$=1+1+1+1=4$
but,
$\displaystyle \sum_{i=1}^{4}a_{i}=(-1)^{1}+(-1)^{2}+(-1)^{3}+(-1)^{4}$
$=-1+1-1+1$
$=0$
The product on the RHS is $0.$
So, for this example, the LHS=4, and the RHS=0.
The statement is not true.
