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Question 619397: I can do this when the numbers are whole, but fractions are really throwing me off. Thank you for any help you can provide!!
Write the geometric series -9/2+3/2-1/2+1/6-...+1/39366 in summation notation. Then using the formula for the sum of a geometric series, compute the sum.
Found 3 solutions by ewatrrr, Theo, KMST:
Answer by ewatrrr(24785)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! the formulas for a geometric progression are:
L = a*r^(n-1)
L = the last term in the sequence.
a = the first term in the sequence.
r = the common ratio.
n = the number of terms in the sequence.
S = a * (1-r^n) / (1-r)
S = the sum of the terms in the sequence.
your sequence is as follows:
term number
1 -9/2
2 3/2
3 -1/2
4 1/6
L 1/39366
The last term in your sequence is equal to 1/39366
your common ratio should be (3/2) / (-9/2)
this is equivalent to:
(3/2) * (2/-9) which is equivalent to:
(3*2) / (2*(-9)) which is equivalent to:
6 / (-18) which is equivalent to:
-(1/3)
if you multiply -9/2 by -1/3, you get 9/6 which is equal to 3/2
if you multiply 3/2 by -1/3, you get -3/6 which is equal to -1/2
if you multiply -1/2 by -1/3, you get 1/6
the common ratio is -1/3 and it is good.
the last term in your sequence is 1/39366
in order to find the sum, you need to know the value of n, unless you are talking about the sum of an infinite series which is not what i understood.
the formula for the last term in the sequence that you showed is:
L = a*r^(n-1)
if you try to solve this by logs, you'll run into difficulty because you can't take the log of a negative number.
in order to use the sum formula, you need to find the value of n and logs aren't any good.
we'll do this by iteration to see if we can come up with the value of n.
let's see how this will work.
we know that the last term in our sequence is equal to 1/39366
we work our way up in the sequence until we get that number and then we'll have the value of n.
we start with the first value of (-9/2)
the common ratio is (-1/3)
formula we'll be using is:
L = (-9/2) * (-1/3) ^ (n-1)
we work our way up as follows:
n-1 L
0 = (-9/2)
1 = (-9/2) * (-1/3) = (-9*-1)/(2*3) = 9/6 = 3/2
...
10 = (-9/2) * (-1/3)^10 = (-9/2) * (-1/59049) = (-9*-1)/(2*59049)
= 9/118098 = 1/13122
11 = (-9/2) * (-1/3)^11 = (-9/2) * (-1/177147) = (-9*-1) / (2*177147)
= 9/354294 = 1/39366 *****
we found the value of n-1 and it is equal to 11.
this means that the value of n is equal to 12.
we have:
n = 12
now we can plug the values into the sum formula to get the sum of the sequence.
the sum of this geometric series is given by the formula:
S = a*(1-r^n)/(1-r)
replacing with known values, we get:
S = (-9/2) * (1 - (-1/3)^12) / (1-(-1/3)
this becomes:
S = (-9/2) * ( 1 - (1/531441) / (4/3) which becomes:
S = (-9/2) * (531440/531441) / (4/3) which becomes:
S = (-9/2) * (531440/531441) * (3/4) which becomes:
S = (-9 * 531440 * 3) / (2 * 531441 * 4) which becomes:
S = -3.374993649
Because this is so messy, I confirmed the results using Excel as shown below:
a = -4.5
r = -0.333333333
formula = a*r^(n-1)
n results of formula for each n
1 -4.5
2 1.5
3 -0.5
4 0.166666667
5 -0.055555556
6 0.018518519
7 -0.00617284
8 0.002057613
9 -0.000685871
10 0.000228624
11 -7.62079E-05
12 2.54026E-05
sum of all results
total -3.374993649
the results are confirmed so the answer looks good.
you caught a bear this time.
not only did they give you off the wall numbers to work with that are exceptionally difficult to do by hand, but they also hit you with a zinger in that you couldn't even use logs to find the value of n because you were dealing with logs of negative numbers which is not allowed.
Answer by KMST(5328)  (Show Source):
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