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

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About mvtrinh

Retired high school math teacher.
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