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\(Compute. i+i^2+i^3+\cdots+i^{258}+i^{259}.\)

 Jul 24, 2016

Best Answer 

 #3
avatar+26367 
+11

Compute.

\(i+i^2+i^3+\cdots+i^{258}+i^{259}.\)

 

geometric series:
\(\begin{array}{|rcll|} \hline a &=& i\\ r &=& i \\ n &=& 259 \\\\ s &=& i\cdot \dfrac{(1-i^{259}) }{1-i}\\\\ s &=& \dfrac{i-i^{260} }{1-i} \qquad & | \qquad i^{4m} &=& 1 \\\\ s &=& \dfrac{i-i^{4\cdot 65} }{1-i} \qquad & | \qquad i^{4\cdot 65} &=& 1 \\\\ s &=& \dfrac{i-1}{1-i} \\\\ s &=& -\dfrac{1-i}{1-i} \\\\ s &=& -1 \\ \hline \end{array} \)

 

laugh

 Jul 25, 2016
 #1
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0

∑[ i^n], for n=1 to 259= -1

 Jul 24, 2016
 #2
avatar+118608 
+6

Compute. i+i^2+i^3+\cdots+i^{258}+i^{259}.

 

\(Compute. i+i^2+i^3+\cdots+i^{258}+i^{259}\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ \mbox{and so the pattern continues, so}\\ i^2+i^4+\dots +i^{256}=0\\ i^1+i^3+\dots +i^{255}=0\\ \mbox{so this leaves}\\ =0+i^{257}+i^{258}+i^{259}\\ =i^{4*64+1}+i^{4*64+2}+i^{4*64+3}\\ =i-1-i\\ =-1 \)

 Jul 25, 2016
 #3
avatar+26367 
+11
Best Answer

Compute.

\(i+i^2+i^3+\cdots+i^{258}+i^{259}.\)

 

geometric series:
\(\begin{array}{|rcll|} \hline a &=& i\\ r &=& i \\ n &=& 259 \\\\ s &=& i\cdot \dfrac{(1-i^{259}) }{1-i}\\\\ s &=& \dfrac{i-i^{260} }{1-i} \qquad & | \qquad i^{4m} &=& 1 \\\\ s &=& \dfrac{i-i^{4\cdot 65} }{1-i} \qquad & | \qquad i^{4\cdot 65} &=& 1 \\\\ s &=& \dfrac{i-1}{1-i} \\\\ s &=& -\dfrac{1-i}{1-i} \\\\ s &=& -1 \\ \hline \end{array} \)

 

laugh

heureka Jul 25, 2016
 #4
avatar+118608 
+5

Thanks Heureka,

I had not thought to do it as a GP  :))

Melody  Jul 25, 2016

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