The value of $sum_(r=0)^50i^((2r+1)!)$ where i is $sqrt(-1)$ ?

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Answer 1 · 2017-06-09T16:55:53

$i+48$

$i^((2r+1)!)=i$ for $r=0$

and

$i^((2r+1)!)=-1$ for $r = 1$

and

$i^((2r+1)!)=1$ for $r > 1$

so

$sum_(r=0)^50i^((2r+1)!) = i-1+49 = i+48$

The value of sum_(r=0)^50i^((2r+1)!)sum_(r=0)^50i^((2r+1)!) where i is sqrt(-1)sqrt(-1) ?