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sigma notation


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#1 Hev

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Posted 28 February 2007 - 08:10 PM

Find Sn= (n-1) above sigma.gif r=0 under 3^r

help me please!
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#2 Steve

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Posted 01 March 2007 - 09:00 PM

OK, first of all just a general thing about the notation. It's nothing special! All \sum_{r=0}^{n-1}3^r means is:



\begin{align*}S_n=\sum_{r=0}^{n-1}3^r = 3^0 + 3^1 + 3^2 +\cdots+3^{n-1}\end{align*}

When the number on top of the \sum is finite (i.e. not \infty) this is called a partial (or finite) sum.

There are two special types of sequences: arithmetic and geometric. For these we have formulae for the nth partial sum.

You can check to see if the terms come from a geometric sequence by dividing any term by the term immediately before it. It is a geometric series if this number is the same no matter which term you do this for. This number is called the common ratio.

In our case, \frac{3^1}{3^0}=\frac{3^2}{3^1}=\frac{3^3}{3^2}=\cdots=\frac{3^{n-1}}{3^{n-2}}=3. So the common ratio is r=3.

The formula for the sum to n terms where the terms come from a geometric sequence is:



\begin{align*}S_n=\sum_{k=0}^{n-1} ar^{k}=\frac{a(1-r^n)}{1-r}\end{align*}

where r is the common ratio and the first term of the sum is a. In our case a=3^0=1.

Does this help?
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#3 Hev

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Posted 01 March 2007 - 11:52 PM

yeah i got it all now thanks smile.gif

think i was just far too tired for maths.

know the feeling? haha
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#4 Steve

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Posted 04 March 2007 - 06:24 PM

Certainly do smile.gif
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