Difference between revisions of "1964 AHSME Problems/Problem 34"
(Created page with "==Problem== If <math>n</math> is a multiple of <math>4</math>, the sum <math>s=1+2i+3i^2+\cdots+(n+1)i^n</math>, where <math>i=\sqrt{-1}</math>, equals: <math>\textbf{(A) }1+i\...") |
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If <math>n</math> is a multiple of <math>4</math>, the sum <math>s=1+2i+3i^2+\cdots+(n+1)i^n</math>, where <math>i=\sqrt{-1}</math>, equals: | If <math>n</math> is a multiple of <math>4</math>, the sum <math>s=1+2i+3i^2+\cdots+(n+1)i^n</math>, where <math>i=\sqrt{-1}</math>, equals: | ||
| − | <math>\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)</math> | + | <math>\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad</math> |
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| + | <math>\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)</math> | ||
Revision as of 18:07, 5 March 2014
Problem
If 
is a multiple of 
, the sum 
, where 
, equals:
![$\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)$](http://latex.artofproblemsolving.com/b/1/8/b188c3a33d644a07c1424325ccc138a7f1c44cd8.png)
