# How do I write sigma notation for a sequence and then find the partial sum?I'm given the sequence: 1, 3, 9, 27, 81, 243... I'm supposed to turn that into sigma notation, and then find the partial...

How do I write sigma notation for a sequence and then find the partial sum?

I'm given the sequence:

1, 3, 9, 27, 81, 243...

I'm supposed to turn that into sigma notation, and then find the partial sum S50, S100 and S200.

I'm completely clueless about how to do this.

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### 3 Answers

If you see the series that you have been given 1, 3, 9, 27, 81, 243..., you see that each term is 3 times the earlier term. This is a geometric progression.

The nth term of a geometric progression can be written as ar^(n-1), where a is the first term, and r is the common ratio.

Also the sum of the first n terms of a geometric series is a(r^(n+1)-1)/(r-1).

Here a = 1 and r = 3. Therefore we can write Tn = 3^(n-1).

Now in sigma notation:

For the first 50 terms you write sigma(n=0 to 50)[3^(n-1)] = 3^51-1 / (3-1) = (3^51 - 1) / 2

For the first 100 terms you write sigma(n=0 to 100)[3^(n-1)] = 3^101-1 / (3-1) = (3^101 - 1) / 2

For the first 200 terms you write sigma(n=0 to 200)[3^(n-1)] = 3^201-1 / (3-1) = (3^201 - 1) / 2

Hi **avenged113**,

I think **justaguide** and **neela** had answered the second part of your question beautifully.

Allow me to answer the first part of your question:

Basically, the series, which you had shown, is a summation series of terms with consecutively increasing powers of 3. I had used a dummy variable r (you can use any other letter for this too. Eg. i or j) to represent the individual powers of 3. The dummy variable and its initial value is written below the Summation Sign (Sigma) and the last value (n) is written above the Sigma.

Click on the following link to see it graphically:

https://docs.google.com/View?id=dgnj78dt_3111jgtsxwg2

(eNotes does not allow me to insert graphic directly. I have used Google docs to get around this problem. )

Hope this helps.

Given sequence is 1,3, 9,27, 81, 243.

We notice that the sequence could also be written as 1, 3, 3^2, 3^3, 3^4,.....

Therefore the nth term of the series is 3^3.

We know that a geometric series 1+x+x^2+x^3+...x^n = (x^n+1-1)/(x-1).

Therefore the sum the sequence for nterms Sn = 1+3+3^2+3^3+...3^3 = Sigma (3n) = (3^(n+1) -1)/(3-1).

Therefore S50 = (3^51 -1)/2 .

S100 = (3^101 -1)/2.

S200 = (3^201 -1)/2.