InterviewSolution
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InterviewSolution
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`sum_(n = 1)^(13)(i^(n)+i^(n+1))=(-1+i), n in N`. |
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Answer» `underset(n = 1)overset(13)sum (i^(n)+i^(n+1))=underset(n = 1)overset(13)sum i^(n)(1+i)=(1+i).underset(n = 1)overset(13)sum i^(n)` `=(1+i).(i+i^(2)+i^(3)+....+i^(13)}` `=(1+i)(a(1-r^(n)))/((1-r))`, where a=i, r=i, and n = 13 `(1+i).(i(1-i^(13)))/((1-i))= i(1+i).((1-i))/((1-i))=(-1 + i)." "[because i^(13) = (i^(13) = (i^(4))^(13) xx i = 1 xx i = i]` |
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