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

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

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

2 Answers

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Answer 1 · 2018-05-12T23:53:57

0.

Explanation:

$i^{14} = i^{2} = - 1$ $i^14=i^2=-1$
$i^{15} = i^{14} \cdot i = - 1 \times i = - i$ $i^15=i^14*i=-1×i=-i$
$i^{16} = i^{15} \cdot i = - i \times i = 1$ $i^16=i^15*i=-i×i=1$
$i^{17} = i^{16} \cdot i = 1 \times i = i$ $i^17=i^16*i=1×i=i$

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

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

Answer 2 · 2018-05-12T23:57:18

Answer A: $0$ $0$

Explanation:

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

$= {(- 1)}^{7} + {(- 1)}^{7} \cdot i + {(- 1)}^{8} + {(- 1)}^{8} \cdot i$ $=(-1)^7+(-1)^7*i+(-1)^8+(-1)^8*i$

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

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

And you'll see they all cancel out.

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

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

2 Answers

Explanation:

Explanation:

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Solve i14+i15+i16+i17=i^14 + i^15 + i^16 + i^17= ?

A) 0 B) 1 C) 2i2i D) 1−i 1 - i E) 2+2i2 + 2i

2 Answers

Explanation:

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Impact of this question

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

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