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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: robo_1337 on June 27, 2013, 05:15:07 pm

Title: Sum of Geometric Sequences
Post by: robo_1337 on June 27, 2013, 05:15:07 pm: Can anyone help me out with this, my teacher hasn't explained it very well and, frankly, I hate the textbooks explanation for things...

thanks,

robo
Title: Re: Sum of Geometric Sequences
Post by: mikehepro on June 27, 2013, 09:28:55 pm: It's applying the equation, what happenes is the next number of the sequence have a ratio to the previous one, R, so if the first term is 3, and there's a ratio of 3, the next number will be 9, the next number after that will be 27. If you want to find the sum of such sequences, say the sum of first 10 terms, you find what a (the first term) is, the r(ratio) and use 10 as n, (the number of terms), then all you have to do is sub these values into the equation, and that's it. Beware that there's an equation for r<1 and there's another one for r>1, they are different.
Title: Re: Sum of Geometric Sequences
Post by: Yendall on June 28, 2013, 10:38:04 am: Been a while since i've done this, but I believe these are the formulae:

$(r\neq1)$ = $a\frac{1-r^n}{1-r}$

$(r=1)$ = $\frac{1}{1-r}$
Title: Re: Sum of Geometric Sequences
Post by: plato on June 28, 2013, 04:04:07 pm: Quote from: Yendall on June 28, 2013, 10:38:04 am
Been a while since i've done this, but I believe these are the formulae:

$(r\neq1)$ = $a\frac{1-r^n}{1-r}$

$(r=1)$ = $\frac{1}{1-r}$

The first formula is correct for the sum of n terms of a geometric sequence.

The second formula should read a/(1-r) rather than 1/(1-r) and is then the sum to infinity of a geometric sequence where 0<r<1.

Edit: Exceptions...
If r>=1, the geometric sequence would be divergent and the infinite sum cannot be found unless the value of a is a fraction.

NOTE: If r is a fraction, difficulties can arise if a is also negative.
For example, if a=-4 and r=0.5, then the second term would be the square root of -4 which is not possible in real numbers.

If a and r are both integers, then the sequence is divegent and a sum to infinity cannot be found, For example, a=2 and r=-2 would give the sequence 2, 1/4, 16, 1/256 ....

If r = 1, then the sequence is again divergent. For example, a=2 and r=1 gives a sequence 2,2,2,2... which has no sum to infinity.

If r = -1, then we get a sequence such as 2, -2, 2, -2, ........ which can give at least two possible answers for an infinite sum:
Either (2-2) + (2-2) + (2-2) ..... = 0
or 2 - (2-2) - (2-2) - (2-2).... = 2

If r < -1 and .........

All such exceptions to the geometric series formulas given in the study design could not be examined but some may find them worthy of exploration.
Title: Re: Sum of Geometric Sequences
Post by: Yendall on June 28, 2013, 07:31:41 pm: Quote from: plato on June 28, 2013, 04:04:07 pm
The first formula is correct for the sum of n terms of a geometric sequence.

The second formula should read a/(1-r) rather than 1/(1-r) and is then the sum to infinity of a geometric sequence where -1<r<0 or 0<r<1.

If r>=1 or r<=-1, the geometric sequence would be divergent and the infinite sum cannot be found.
Ah right, sorry I must've missed something. Thanks for clearing that up for Robo
Title: Re: Sum of Geometric Sequences
Post by: robo_1337 on July 01, 2013, 02:18:17 pm: Is there really any difference between the formulas?
The ratio appears to be the same for both $(r\neq1)$ = $a\frac{1-r^n}{1-r}$ and $(r\neq1)$ = $a\frac{r^n-1}{r-1}$ .
ie. $\frac{1-2^2}{1-2}$ = $\frac{2^2-1}{2-1}$ = $3$

robo
Title: Re: Sum of Geometric Sequences
Post by: Professor Polonsky on July 01, 2013, 02:35:05 pm: No, it's the same thing.

$S = a\frac{r^n -1}{r-1} = a\frac{1-r^n}{1-r}$

The latter is probably easier to use when $-1<r<1$

There's also the sum to infinity, $S_{\infty}=\frac{a}{1-r}$ which only applies when $-1<r<1$