Solve $i^14 + i^15 + i^16 + i^17=$ ?

Question

A) 0
B) 1
C) $2i$
D) $1 - i$
E) $2 + 2i$

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Answer 1 · 2018-05-12T23:53:57

0.

$i^14=i^2=-1$
$i^15=i^14*i=-1×i=-i$
$i^16=i^15*i=-i×i=1$
$i^17=i^16*i=1×i=i$

Therefore, $-1-i+1+i=0$

$"So option A is correct."$
$"Generally sum of any four consecutive powers of i is 0"$ .

Answer 2 · 2018-05-12T23:57:18

Answer A: $0$

Every $i^2=-1$
So we rewrite, taking out the $i$ 's two by two:

$=(-1)^7+(-1)^7*i+(-1)^8+(-1)^8*i$

Every even power of $-1$ will give $+1$ , and every odd power will give $-1$ . Rewrite again:

$=-1+(-1)*i+1+1*i$

And you'll see they all cancel out.