Difference between revisions of "Imaginary unit/Introductory"
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==Problem== | ==Problem== | ||
| − | Find the sum of <math>i^1+i^2+\ldots+i^{2006} | + | Find the sum of <math>i^1+i^2+\ldots+i^{2006}</math> |
== Solution == | == Solution == | ||
| − | Let's begin by computing powers of < | + | Let's begin by computing powers of <math>i</math>. |
| − | #< | + | #<math>i^1=\sqrt{-1}</math> |
| − | #< | + | #<math>i^2=\sqrt{-1}\cdot\sqrt{-1}=-1</math> |
| − | #< | + | #<math>i^3=-1\cdot i=-i</math> |
| − | #< | + | #<math>i^4=-i\cdot i=-i^2=-(-1)=1</math> |
| − | #< | + | #<math>i^5=1\cdot i=i</math> |
| − | We can now stop because we have come back to our original term. This means that the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences < | + | We can now stop because we have come back to our original term. This means that the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences <math>i^1+i^2+\ldots+i^{4k}</math> have a sum of zero (k is a natural number). Since <math>2006=4\cdot501+2</math>, the original series sums to the first two terms of the powers of i, which equals <math>-1+i</math>. |
Revision as of 13:38, 26 October 2007
Problem
Find the sum of 
Solution
Let's begin by computing powers of 
.
We can now stop because we have come back to our original term. This means that the sequence i, -1, -i, 1 repeats. Note that this sums to 0. That means that all sequences 
have a sum of zero (k is a natural number). Since 
, the original series sums to the first two terms of the powers of i, which equals 
.
