sum of infinite series
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The above means "(t)(n) = (-2)<sup>1</sup> - n". But I think you mean "t[n] = (-2)^(1 - n)"; that is, t<sub>n</sub> = (-2)<sup>1 - n</sup>.xc630 said:
Please reply with confirmation or correction. When you reply, please include your listing of the first few terms of the sequence, showing the steps you used to try to use these terms to find the common ratio.
Thank you.
Eliz.
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Edit: Ne'mind: The answer is below.
Hello, xc630!
Well, the first term is: \(\displaystyle \,t_{_1}\;=\;(-2)^{1-1} = (-2)^0\;=\;1\)
The second term is: \(\displaystyle \,t_{_2}\;=\;(-2)^{1-2}\;=\;(-2)^{-1}\;=\;-\frac{1}{2}\)
The common ratio is: \(\displaystyle \L\:r\;=\;\frac{t_{_2}}{t_{_1}}\;=\;\frac{-\frac{1}{2}}{1} \;=\;-\frac{1}{2}\)
\(\displaystyle t_1\,+\,t_2\,+\,\cdots\, +\,t_n\,+\,\cdots,\;\) where \(\displaystyle t_n\:=\-2)^{1-n}\)
I need to find the sum of the series but I'm not sure how to find the common ratio
Well, the first term is: \(\displaystyle \,t_{_1}\;=\;(-2)^{1-1} = (-2)^0\;=\;1\)
The second term is: \(\displaystyle \,t_{_2}\;=\;(-2)^{1-2}\;=\;(-2)^{-1}\;=\;-\frac{1}{2}\)
The common ratio is: \(\displaystyle \L\:r\;=\;\frac{t_{_2}}{t_{_1}}\;=\;\frac{-\frac{1}{2}}{1} \;=\;-\frac{1}{2}\)