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
285
4
avatar+603 

 

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

 Apr 9, 2018
 #1
avatar
+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
avatar+10357 
+1

∑[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
avatar+428 
+1

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

 Apr 9, 2018
 #4
avatar+22174 
+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

 Apr 10, 2018

24 Online Users

avatar
avatar
avatar