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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 = 3sum_{n=-42}^5 x^n = 3sum_{n=-5}^42x^(-n) = 3sum_{n=-5}^42(\frac{1}{x})^n $
Answer 3
write it here.
