Question

solve 3(-⅓)^4

Answer 1

Do we have to use Real Numbers? That could be a problem.

Answer 2

@tkhunny well, the original question says to find the sum of the serious.
5
∑ 3(-⅓)^n-1

Answer 3

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.

Answer 4

Are we summing from 1 through 5? Can you use the Equation Editor to provide a better understanding of what is wanted?

Answer 5

\[\sum_{i=1}^{5} 3(-⅓)^n-1\]

Answer 6

does that help @tkhunny

Answer 7

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?

Answer 8

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

Answer 9

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?

Answer 10

Geometric? @tkhunny

Answer 11

Arithmetic!

Answer 12

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

Answer 13

Ohh, okay that makes sense. @tkhunny

Answer 14

Woo-hoo!! What do you get?

Answer 15

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.