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

 Sep 5, 2018
edited by Guest  Sep 6, 2018

Best Answer 

 #2
avatar+98196 
+2

Note the repeating pattern  of  length  4

i^1   = i

i^2  = -1

i^3  = -i

i^4  =  1

 

So....the sum of this series  from i^1  to  i*96   = 0

 

So...the sum of the last  three terms   = i^97  + i^98  + i^99  = i + -1 +  -i   = -1

 

So...this series sums  to  0 + (-1)  = -1

 

 

cool cool cool

 Sep 6, 2018
 #1
avatar+4471 
+1

\(\displaystyle \sum \limits_{k=1}^n~\alpha^k =\dfrac{\alpha \left(\alpha ^n-1\right)}{\alpha -1}\)

 

letting \(\alpha=i,~n=99,~\text{and assuming }i=\sqrt{-1}\)

 

\(\displaystyle \sum \limits_{k=1}^{99} ~i^k = \dfrac{i(i^{100}-1)}{i-1} = 0\)

.
 Sep 6, 2018
 #2
avatar+98196 
+2
Best Answer

Note the repeating pattern  of  length  4

i^1   = i

i^2  = -1

i^3  = -i

i^4  =  1

 

So....the sum of this series  from i^1  to  i*96   = 0

 

So...the sum of the last  three terms   = i^97  + i^98  + i^99  = i + -1 +  -i   = -1

 

So...this series sums  to  0 + (-1)  = -1

 

 

cool cool cool

CPhill Sep 6, 2018
 #3
avatar+21860 
+8

Simplify

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

 

Geometric sequence:

\(\begin{array}{|rcll|} \hline s &=& i^1+&i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99} \quad & | \quad \cdot i \\ i\cdot s &=& &i^2+i^3+i^4+\cdots+ i^{98} + i^{99}+i^{100} \\\\ \hline s-i\cdot s &=& i^1 - i^{100} \\ s(1-i) &=& i^1 - i^{100} \\\\ \mathbf{s} &\mathbf{=} & \mathbf{ \dfrac{i - i^{100}}{1-i} } \quad & | \quad i^{100}= \left(i^2 \right)^{50}=(-1)^{50}=1 \\\\ s & = & \dfrac{i-1}{1-i} \\ s & = & -\left( \dfrac{1-i}{1-i} \right) \\ \mathbf{s} &\mathbf{=} & \mathbf{-1} \\ \hline \end{array}\)

 

\(\boxed{\large{i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99} = -1}}\)

 

laugh

 Sep 6, 2018

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