+0  
 
0
40
4
avatar+414 

 

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

gueesstt  Apr 9, 2018
Sort: 

4+0 Answers

 #1
avatar
+1

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

Guest Apr 9, 2018
edited by Guest  Apr 9, 2018
 #2
avatar+9208 
+1

∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i + 1....................- i + 1 + i - 1 - i = - 1

 

Omi67  Apr 9, 2018
edited by Omi67  Apr 9, 2018
 #3
avatar+418 
+1

∑[i^n, n, 1, 99] =i - 1 - i + 1 + i - 1 - i + 1....................- i + 1 + i - 1 - i = - 1

jakesplace  Apr 9, 2018
 #4
avatar+19207 
+1

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

 

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

 

laugh

heureka  Apr 10, 2018

31 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details