Question 117202: i+i^2+i^3+...+i^49=
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Given the problem:
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i+i^2+i^3+...+i^49= ???
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One way you can do this is by "patterns" ...
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Note that the first four terms of this series is:
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i^1 = +i
i^2 = -1
i^3 = -i
i^4 = +1
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When you add these four terms, they combine to equal zero (+1 - 1 + i - i = 0
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Now look at the next four terms in the series:
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i^5 = i^4 * i = +1*i = +i
i^6 = i^4 * i * i = 1 * i * i = i^2 = -1
i^7 = i^4 * i^3 = +1 * -i = -i
i^8 = i^4 * i^4 = +1 * +1 = +1
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This shows you that the sum in this series is +i - 1 - i + 1 and is the same as the series of
the first four terms. Therefore, it also totals to zero.
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If you do the next 4 terms in the series ... i^9 through i^12 you will find that it also produces
the four terms +i - 1 - i + 1 and therefore also sums to zero.
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The pattern cycles through each group of 4 terms ... each group of 4 adding to zero. This means
that up through i^48 you have 12 groups of 4 terms ... each group equaling zero. Therefore,
the sum of the series i + i^2 + i^3 + i^4 + i^5 .... + i^45 + i^46 + i^47 + i^48 totals zero.
So the sum of the series [i + i^2 + i^3 + i^4 + i^5 .... i^45 + i^46 + i^47 + i^48] + i^49
is equal to zero + i^49, and since i^49 is the the first term in the next group of 4, it
should have a value of i. Think of it this way also ... i^49 = i^48 * i = (i^4)^12 * i
and i^4 = +1. So (i^4)^12 is (+1)^12 = 1. So (i^48)*i = 1*i = i.
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Putting it all together .... the sum of this entire series is 0 for the first 48 terms and + i for
the 49th term. Therefore, through the 49 terms of the series the answer to this problem is
just ... i.
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Hopefully this all makes sense when you carefully work your way through it.
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