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# Help!!!!!

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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

#2
+100588
+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

Sep 6, 2018

#1
+5097
+1

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

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

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

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Sep 6, 2018
#2
+100588
+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

CPhill Sep 6, 2018
#3
+22198
+8

Simplify

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

Geometric sequence:

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

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

Sep 6, 2018