Sum of 1/i

Find the sum of
$\dfrac{1}{i}+\dfrac{3}{i^3}+\dfrac{5}{i^5}+\dfrac{7}{i^7}+\dfrac{9}{i^9}+\cdots+\dfrac{53}{i^{53}}$
where $i=\sqrt{\textrm{-}1}$.
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
$S=\dfrac{1}{i}+\dfrac{3}{i^3}+\dfrac{5}{i^5}+\dfrac{7}{i^7}+\dfrac{9}{i^9}+\cdots+\dfrac{53}{i^{53}}$
Multiply both sides by $1/i$
$\dfrac{1}{i}S=\dfrac{1}{i^2}+\dfrac{3}{i^4}+\dfrac{5}{i^6}+\dfrac{7}{i^8}+\dfrac{9}{i^{10}}+\cdots+\dfrac{53}{i^{54}}$

$=\dfrac{1}{\textrm{-}1}+\dfrac{3}{1}+\dfrac{5}{\textrm{-}1}+\dfrac{7}{1}+\dfrac{9}{\textrm{-}1}+\cdots+\dfrac{51}{1}+\dfrac{53}{\textrm{-}1}$
$=-1+3-5+7-9+\cdots+51-53$
The above expression has $27$ terms with the middle term equal $27$
$-1+3-5+7-\cdots-21+23-25+\underline{27}-29+31-33+35-\cdots-49+51-53$
We combine the first term to the last term, the second to the next to last and so on
$(-1-53)+(3+51)+(-5-49)+(7+47)+\cdots+(23+31)+(-25-29)+27$
We end up with seven $-54$ terms, six $+54$ terms and the middle term $27$. The six $-54$ terms cancel the six $+54$ terms leaving the remaining two terms $-54+27=-27$.
$\dfrac{1}{i}S=-27$
$S=-27i$

Answer: $-27i$
Alternative solution
$-1+3-5+7-\cdots-21+23-25+\underline{27}-29+31-33+35-\cdots-49+51-53$
We combine the terms in pairs
$(-1+3)+(-5+7)+(-9+11)+\cdots+(-45+47)+(-49+51)-53$
We end up with thirteen $\textrm{+}2$ terms and the remaining term $-53$
$13(2)-53=-27$