Question 29589: I am having trouble with this problem, could you please help me.
3) Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following.
PROBLEMS:
b) Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 terms?
c) Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms?
d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?
Thank you in advance for all your help.
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website!
Use the geometric sequence of numbers 1, 1/3, 1/9 , 1/27… to find the following.
PROBLEMS:
b) Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 terms?
c) Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms?
d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?
The general geometric progression is
a, ar,ar^2,ar^3,ar^4, .....,ar^(n-1),......
The given geometric progression is 1, 1/3, 1/9 , 1/27…....
So here a = 1 and r = (1/3)
The formula for the sum to n terms is
S = a[1-(r^n)]/(1-r) ----(I)
1)To find the sum of the first 10 terms of 1, 1/3, 1/9 , 1/27…....
Putting a = 1, r= (1/3) and n =10
S= a/(1-r)[1-(r^n)]
=1/[1-(1/3)]X[1-(1/3)10]
=[1/(2/3)[1-1/(3^10)] (using (u/v)^p = (u^p)/(v^p) and here u = 1and v = 3 and p= 10 )
=(3/2)[3^(10) - 1]/(3^10)
= [3^(10) - 1]/[2X(3^9)] (cancelling 3 in the nr and in the dr)
2)To find the sum of the first 12 terms of 1, 1/3, 1/9 , 1/27…....
Putting a = 1, r= (1/3) and n =12
S= a/(1-r)[1-(r^n)]
=1/[1-(1/3)]X[1-(1/3)12]
=[1/(2/3)[1-1/(3^12)] (using (u/v)^p = (u^p)/(v^p) and here u = 1and v = 3 and p= 12 )
=(3/2)[3^(12) - 1]/(3^12)
= [3^(12) - 1]/[2X(3^11)] (cancelling 3 in the nr and in the dr)
d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?
If the number n keeps increasing and for r, the common ratio < 1
as in this example r = (1/3) which is less than 1
r^n keeps decreasing until fianlly when n is very large,so large that it tends to infinity,
then (r^n) tends to zero which implies [1-(r^n] tends to (1-0) = 1 ----(*)
And the sum to infinity of this above series 1+1/3+1/9 +1/27…...
S = a[1-(r^n)]/(1-r) ----(I)
becomes S = aX(1)/(1-r)
= 1/(1-r)
= 1/[1-(1/3)]
= 1/(2/3)
=3/2
Note: I don't follow exactly this statement:what number does it appear that the sum will always be smaller than?
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