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simplify i^1+i^2+i^3+ ... + i^ 97 + i^ 98 +i^ 99

 Sep 8, 2018
 #1
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+1

hint: remember the formula for a sum of a geometric sequence

 Sep 8, 2018
 #2
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+1

∑[ i^n, n, 1, 99]=-1, because[i^1 + i^2 +i^3 + i^4]=0 + [i^5+i^6+i^7+i^8]=0+0=0. This pattern continues until i^96, which is a multiple of 4. i^97=i, i^98=-1, i^99=-i. So add the last 3 together:

i + (-1) + (-i) = -1 - the two "i" cancel leaving -1 as the answer.

 Sep 8, 2018

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