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Compute \(1+i+i^2+i^3+i^4+\cdots+i^{2009}\)

michaelcai  Dec 12, 2017
 #1
avatar+86888 
+1

Note

 

i1 + 4n + i 2 + 4n + i3 + 4n  + i 4 + 4n  =   0     where n  is an integer  ≥ 0

 

So.....  we have

 

1  +  0  + i 2009 =

 

1 +  i 1 + 4(502)  = 

 

1 +  i * i2008 =

 

1 +  i *  ( i2 )1004

 

1 + i  (-1)1004      { (-1)2n   = 1 }

 

1  +  i

 

 

cool cool cool

CPhill  Dec 12, 2017
 #3
avatar+19482 
+2

Compute $1+i+i^2+i^3+i^4+\cdots+i^{2009}$.

 

Compute \(1+i+i^2+i^3+i^4+\cdots+i^{2009}.\)

 

geometric sequence: \(a = 1,~ r = i\)

The sum is \(\begin{array}{|rcll|} \hline s &=& 1\cdot \dfrac{i^{2010}-1}{i-1} \\ \hline \end{array}\)

 

\(\begin{array}{|rclcrcl|} \hline s &=& 1\cdot \left( \dfrac{i^{2010}-1}{i-1} \right) \\ && & i^{2010} &=& i^{2\cdot1005} \\ && & &=& (i^2)^{1005} \quad & | \quad i^2 = -1 \\ && & &=& (-1)^{1005} \\ && & &=& -1 \\ s &=& \left(\dfrac{-1-1}{i-1}\right)\cdot \left(\dfrac{i+1}{i+1}\right) \\\\ &=& \dfrac{(-1-1)(i+1)}{(i-1)(i+1)} \\\\ &=& \dfrac{(-1-1)(i+1)}{(i^2-1)} \quad & | \quad i^2 = -1 \\\\ &=& \dfrac{(-1-1)(i+1)}{(-1-1)} \\\\ &=& i+1 \\ \hline \end{array}\)

 

\( \mathbf{1+i+i^2+i^3+i^4+\cdots+i^{2009} = 1+i} \)

 

 

laugh

heureka  Dec 12, 2017

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