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Simplify \[i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\]

 
Guest Jan 11, 2018
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 #1
avatar+80793 
+1

 \( \[i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\]\)

 

Note that

 

i1  = i

i2  = -1

i3  =  i * i2  =  i * -1  =  - i

i4  =  i2 * i2  =  -1 * - 1  =  1

 

So....the sum of these  = 0

 

And this pattern continues for each partial sum of

i1+4n, i2+4n, i3 + 4n and i4 + 4n       where  n is an integer ≥ 0

 

So......we  will have

 

0  +  0  + 0  +  .......+  0  + 0  +  i97 + i98 + i99  =

 

i1+ 4(24)  +  i 2 + 4(24)  + i 3 + 4(24)   =

 

[    i    ]     +  [    -1   ]        +    [  - i  ]    =

 

 

-1

 

 

 

cool cool cool

 
CPhill  Jan 11, 2018
 #3
avatar+18827 
+1

Can someone help me I do not understand the question

Simplify \(i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\)

 

Geometric sequence ratio = i

\(\begin{array}{|rcll|} \hline s_{99} &=& \frac{ i - i^{100} } {1-i} \qquad & | \qquad i^{100} = i^{2\cdot 50} = (i^2)^{50} = (-1)^{50} = 1 \\\\ s_{99} &=& \frac{ i - 1 } {1-i} \\\\ s_{99} &=& - \left( \frac{ 1 - i } {1-i} \right) \\\\ s_{99} &=& - 1 \\ \hline \end{array} \)

 

laugh

 
heureka  Jan 11, 2018

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