InterviewSolution
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InterviewSolution
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Show that the sum `(1 + i^(2) + i^(4)+....+i^(2n))` is 0 when n is odd and 1 when n is even. |
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Answer» Let `S = 1 + i^(2) + i^(4) +....+i^(2n)`. This is clearly a GP having (n + 1) terms with a = 1 and r = `i^(2)` = -1. `therefore" "S = (a(1-r^(n+1)))/((1+r))=(1xx{1-(i^(2))^(n+1)})/((1-i^(2)))` `" "= ({1-(-1)^(n+1)})/(1-(-1))=({q-(-1)^(n+1)})/(2)` `" "={{:((1)/(2)(1-1)=0",""when n is odd"),((1)/(2)(1+1)=1",""when n is even".):}` |
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