# Doubt from series sum

Ans??? @Achyut_2020

Bla bla

Is it -49?

50?

I will be considering a = e^{\frac{2\pi}{101}} It doesn't matter which root of unity I take, it'll be a symmetric sum so I chose this one arbitrarily.

\sum_{i=1}^{100} \frac{a^{3i}}{1+a^i}

= \sum_{i=1}^{100} (a^{2i} - a^{i} + 1-\frac{1}{1+a^i})

= \sum_{i=1}^{100} a^{2i} =\sum_{i=1}^{100} e^{2*2i\pi/101} = -1 -> Try proving/disproving this

= \sum_{i=1}^{100} -a^{i} =\sum_{i=1}^{100}- e^{i\pi/100} = 1

= \sum_{i=1}^{100} -\frac{1}{1+a^i} = 50 -> To find this sum:

Consider \frac{z^{101}-1}{z-1} = \Pi_(z-\alpha^{i}) //Where alpha denotes first 101th complex root of unity.
Take ln on both sides and differentiate, then proceed to put z = -1

Adding the sums, we get 150. (did it fast so might have made a calculation error.)

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Getting 50 @Achyut_2020 ?

Ans is given 50