Called Complex For A(nother) Reason

Algebra Level 3

For $n \geq 4$ , evaluate

$\large i^{0!}+i^{1!}+i^{2!}+i^{3!}+\cdots+i^{n!}.$

Clarification :

$i = \sqrt{-1}$ .
$n! = n\times(n-1)\times(n-2)\times\dots\times3\times2\times1$ .

Inspired by the series Called Complex For A Reason by Zandra Vinegar. Try her Part I , Part II and Part III

0

n-3

2i

(n-5)+2i

1 solution

Kay Xspre
Feb 24, 2016

There are many ways to approach this, but first, expand the exponents. The exponents will be $j! = 1, 1, 2, 6, 24, \dots, n!\:; j\:\in \{0, 1, 2, \dots, n\}$ Here, we can notice that from $4! = 24$ onward, the exponent will be multiple of 4, causing $i^{n!} = 1$ for $n \geq 4$ . For others, we can simply find $i^{j!}$ as $i^{j!} = i, i, -1, -1, 1, 1, \dots, 1\:; j\:\in \{0, 1, 2, \dots, n\}$ We count that there will be $(n-4)+1$ terms of $1$ and two terms each of $-1$ and $i$ , therefore the sum is $((n-4)+1)+2(-1+i) = (n-5)+2i$

Called Complex For A(nother) Reason

Inspired by the series Called Complex For A Reason by Zandra Vinegar. Try her Part I , Part II and Part III

1 solution

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