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

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

michaelcai  Dec 12, 2017
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#1
+81023
+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

CPhill  Dec 12, 2017
#3
+18829
+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 $$\begin{array}{|rcll|} \hline s &=& 1\cdot \dfrac{i^{2010}-1}{i-1} \\ \hline \end{array}$$

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

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

heureka  Dec 12, 2017

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