Revision as of 17:10, 6 September 2011

Simplify this summation

$ \sum_{n=-42}^5 3^{n+1} (1+j)^n  $

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Answer 1

TA's comments: Any complex number can be written as one single complex exponential. i.e. $ a+jb=\sqrt{a^2+b^2}e^{j\theta}, where\ tan\theta = \frac{b}{a} $

Answer 2

Set $ x=3+j3 $. Note that $ |x|>1 $.

$ \sum_{n=-42}^5 3^{n+1} (1+j)^n = 3\sum_{n=-42}^5 x^n = 3\sum_{n=-5}^{42}x^{-n} = 3\sum_{n=-5}^{42}(\frac{1}{x})^n  $
$  = 3(\sum_{n=-5}^{\infty}(\frac{1}{x})^n - \sum_{n=43}^{\infty}(\frac{1}{x})^n) = 3(\frac{(\frac{1}{x})^{-5}}{1-\frac{1}{x}} - \frac{(\frac{1}{x})^{43}}{1-\frac{1}{x}}) = 3(\frac{x^6-x^{-42}}{x-1}) = -4037-j2692  $

Instructor's comments: There is a much shorter solution using the finite geometric series formula. Note that, when the sum is finite, one does not have to worry about convergence. In particular, the formula works even if the norm of the argument is greater than one. -pm

Answer 3

$ \sum_{n=-42}^5 3^{n+1} (1+j)^n = \sum_{n=-42}^5 3^{n+1} (\sqrt{2} e^{j\pi/4})^n $

By letting l = n+42,

$ \sum_{l=0}^{47} 3^{l-41} (\sqrt{2} e^{j\pi/4})^{l-42} =        3^{-41}(\sqrt{2}e^{j\pi/4})^{-42}\sum_{l=0}^{47} (3\sqrt{2}e^{j\pi/4})^l =         \frac{1 - (3\sqrt{2}e^{j\pi/4})^{48}}{1 - 3\sqrt{2}e^{j\pi/4}}3^{-41}(\sqrt{2}e^{j\pi/4})^{-42}   $

This is as far as I could go trying to type these long equations...

Answer 4

use finite geometric series formula $ S=\frac{a_1(1-r^n)}{1-r};a1=3,r=(3+3j)^(-42),n=47,then S=-4037-j2692 $

Answer 5

$ \sum_{k=0}^{n-1}ar^k=a\frac{1-r^n}{1-r} $

$ \begin{align} \sum_{n=-42}^5 3^{n+1} (1+j)^n &= 3\sum_{n=-42}^5 (3(1+j))^n \\ &=3\frac{1-(3(1+j))^6}{1-3(1+j)}+3\frac{1-(3(1+j))^{-41}}{1-(3(1+j))^{-1}}-3 \end{align} $

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