Circular sums (II)

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∑ n = 0 20 e n i π / 2 + l n ( 2 ) n = a + b i \large \sum_{n=0}^{20} e^{n i \pi /2+ln(2)n}= a+bi

If the relationship above holds true for real numbers a a and b b , calculate a + b a+b .


The answer is 419431.

William Whitehouse by William Whitehouse , 22, United Kingdom

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1 solution

The first step is to notice that this is the sum of a geometric progression, with r=2i.

With this in mind the sum can be evaluated using the sum of a geometric progression formula S n = a 1 ( 1 − r n + 1 ) 1 − r S_{n} = \frac{a_{1}(1-r^{n+1})}{1-r}

with n=20 , a=1 this evaluates to 838861-419430i so a=838861 b=-419430 and a+b=419431

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