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

 Apr 28, 2019
 #1
avatar+2547 
+2

Imaginary number rules,

 

To the power of -

 

1: i

2: -1

3: -i

4: 1

5: i

 

and so on.

-------------------------------

Now we have to find how many times each will occur throughout the sequence.

 

Since 96th term will be 1, this means that each power of i outcome will occur 24 times. So we have 24i-24-24i+24i

 

97th term will be i

 

98th term will be -1

 

99th term will be -i

 

So add the three above to the expression we had earlier to get : 25i-25-25i+24i

 

Simplifies to -25+24i

 

\(\boxed{-25+24i}\)

.
 Apr 28, 2019
edited by CalculatorUser  Apr 28, 2019
 #2
avatar+23903 
+2

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

 \(\begin{array}{|rcll|} \hline && \displaystyle i^1+i^2+i^3+i^4+i^5+\cdots+ i^{97} + i^{98}+i^{99} \\\\ &=& (i^1+i^2+i^3+i^4)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& (i^1+i^2)(1+i^2)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \quad | \quad i^2 = -1 \\\\ &=& (i^1+i^2)(1-1)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& (i^1+i^2)(\underbrace{1-1}_{=0})(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& 0-i^{100} \\\\ &=& -(i^2)^{50} \quad | \quad i^2 = -1 \\\\ &=& -(-1)^{50} \\\\ &=& -1 \\ \hline \end{array}\)

 

laugh

 Apr 29, 2019
edited by heureka  Apr 29, 2019

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