We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
337
3
avatar

Simplify \[i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.\]

 Jan 11, 2018
 #1
avatar+101149 
+1

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

 

Note that

 

i1  = i

i2  = -1

i3  =  i * i2  =  i * -1  =  - i

i4  =  i2 * i2  =  -1 * - 1  =  1

 

So....the sum of these  = 0

 

And this pattern continues for each partial sum of

i1+4n, i2+4n, i3 + 4n and i4 + 4n       where  n is an integer ≥ 0

 

So......we  will have

 

0  +  0  + 0  +  .......+  0  + 0  +  i97 + i98 + i99  =

 

i1+ 4(24)  +  i 2 + 4(24)  + i 3 + 4(24)   =

 

[    i    ]     +  [    -1   ]        +    [  - i  ]    =

 

 

-1

 

 

 

cool cool cool

 Jan 11, 2018
 #3
avatar+22290 
+1

Can someone help me I do not understand the question

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

 

Geometric sequence ratio = i

\(\begin{array}{|rcll|} \hline s_{99} &=& \frac{ i - i^{100} } {1-i} \qquad & | \qquad i^{100} = i^{2\cdot 50} = (i^2)^{50} = (-1)^{50} = 1 \\\\ s_{99} &=& \frac{ i - 1 } {1-i} \\\\ s_{99} &=& - \left( \frac{ 1 - i } {1-i} \right) \\\\ s_{99} &=& - 1 \\ \hline \end{array} \)

 

laugh

 Jan 11, 2018

21 Online Users

avatar