anonymous
  • anonymous
solve 3(-⅓)^4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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tkhunny
  • tkhunny
Do we have to use Real Numbers? That could be a problem.
anonymous
  • anonymous
@tkhunny well, the original question says to find the sum of the serious. 5 ∑ 3(-⅓)^n-1
tkhunny
  • tkhunny
Aha!! Please provide the entire question int he future. Also, provide your best efforts at a solution. Does that appear to be a type of series for which the sum can be easily computed? Generally, only Arithmetic and Geometric fall into this calssification.

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tkhunny
  • tkhunny
Are we summing from 1 through 5? Can you use the Equation Editor to provide a better understanding of what is wanted?
anonymous
  • anonymous
\[\sum_{i=1}^{5} 3(-⅓)^n-1\]
anonymous
  • anonymous
does that help @tkhunny
tkhunny
  • tkhunny
That is SWEET!!!!!! Excellent work. Okay, now answer my question. Can you identify the TYPE of series? If you have hope of finding the sum, it had better be Arithmetic or Geometric. Is it either of those?
anonymous
  • anonymous
It doesn't say on the paper. Usually when I do these, i'm not sure what the series is until i have solved it. I'm not sure what to classify it as though @tkhunny
tkhunny
  • tkhunny
Then you need to learn to classify them. Arithmetic - Successive terms differ by a fixed constant. Example: 1, 2, 3, 4, ... Each term is one (1) more than it's predecessor. Geometric - Successive terms differ by a fixed ratio. Example: 1, 2, 4, 8, ... Each term is twice it's predecessor. I ask again, still with a big smile, what kind of series have we?
anonymous
  • anonymous
Geometric? @tkhunny
anonymous
  • anonymous
Arithmetic!
tkhunny
  • tkhunny
That's it. It is then trivial to add it up. Usually, it's: The 1st Term: a = 3 The common ratio: r = -1/3 The number of terms: 5
anonymous
  • anonymous
Ohh, okay that makes sense. @tkhunny
tkhunny
  • tkhunny
Woo-hoo!! What do you get?
tkhunny
  • tkhunny
Just for the interested reader, it was supposed to be \((-1/3)^{n-1}\). So it actually is a geometric series. \((-1/3)^{n} - 1\) would not be a geometric series.

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