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.