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

Apr 9, 2018

#1
+1

∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i.................... 1 + i - 1 - i = - 1. Sorry for the typo!!.

Apr 9, 2018
edited by Guest  Apr 9, 2018
#2
+10357
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∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i + 1....................- i + 1 + i - 1 - i = - 1

Apr 9, 2018
edited by Omi67  Apr 9, 2018
#3
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∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i + 1....................- i + 1 + i - 1 - i = - 1

Apr 9, 2018
#4
+22174
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Simplify $i^1+i^2+i^3+\cdots+ i^{97} + i^{98}+i^{99}.$

$$\begin{array}{|rcll|} \hline && \displaystyle i^1+i^2+i^3+i^4+i^5+\cdots+ i^{97} + i^{98}+i^{99} \\\\ &=& (i^1+i^2+i^3+i^4)(1+i^4+i^8+i^{12}+\cdots+ i^{96} )-i^{100} \\\\ &=& (i^1+i^2)(1+i^2)\sum \limits_{k=0}^{24} (i^{4k})-i^{100} \quad & | \quad i^2 = -1 \\\\ &=& (i^1+i^2)(\underbrace{1+i^2}_{=0})\sum \limits_{k=0}^{24} (i^{4k})-i^{100} \\\\ &=& 0-i^{100} \\\\ &=& -(i^2)^{50} \\\\ &=& -(-1)^{50} \\\\ &=& -1 \\ \hline \end{array}$$

Apr 10, 2018