+0

# Can someone help me I do not understand the question

0
337
3

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

Jan 11, 2018

#1
+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

Jan 11, 2018
#3
+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

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

Jan 11, 2018