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

A) 0
B) 1
C) 2i2i
D) 1 - i1i
E) 2 + 2i2+2i

2 Answers

0.

Explanation:

i^14=i^2=-1i14=i2=1
i^15=i^14*i=-1×i=-ii15=i14i=1×i=i
i^16=i^15*i=-i×i=1i16=i15i=i×i=1
i^17=i^16*i=1×i=ii17=i16i=1×i=i

Therefore, -1-i+1+i=01i+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.

May 12, 2018

Answer A: 00

Explanation:

Every i^2=-1i2=1
So we rewrite, taking out the ii's two by two:

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

Every even power of -11 will give +1+1, and every odd power will give -11. Rewrite again:

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

And you'll see they all cancel out.