1/9 , 1/27… to find the following: Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 and 12 terms? Can you give me the formula or the example because I am just really lost.
2006-11-02 05:11:00 · 4 answers · asked by Rhonda O 1 in Science & Mathematics ➔ Mathematics
the sum of a series of n terms...of a geometric progressive series...is given by
Sum =t1(r^n-1)/(r-1)
now substituting the value we get for 10 terms
s10=1(1-(1/3)^10)/(1-(1/3))
s13=1(1-(1/3)^13)/(1-(1/3))
similarly for n terms sn=1(1-(1/3)^n)/(1-(1/3))
2006-11-02 05:35:55 · answer #1 · answered by indurti karthik 2 · 0⤊ 0⤋
Given the geometric sequence
a_n = r^n, n = 0, 1, 2, 3, ...
(in your case, r=1/3, so you get (1/3)^0 = 1, (1/3)^1 = 1, (1/3)^2 = 1/9, etc.)
the formula for the sum of the first n terms is
(r^n - 1) / (r - 1)
For example, the first 4 terms with r=1/3 are 1,1/3,1/9,1/27. Therefore, the sum is
((1/3)^4 - 1) / (1/3 - 1) = (1/81 - 1) / (1/3 - 1)
= (-80/81) / (-2/3) = (80/81)*(3/2) = 40/27, which is what you would get from computing 1 + 1/3 + 1/9 + 1/27.
2006-11-02 05:26:38 · answer #2 · answered by James L 5 · 0⤊ 0⤋
It appears that each number in the sequence is divided by 3 to get the number that follows it. Thus the sequence for the first 12 numbers is 1, 1/3, 1/9, 1/27, 1/81, 1/243, 1/729, 1/2187, 1/6561, 1/19683, 1/59049, and 1/177147. The sum of the first 10 terms would be 9841/6561 or reduced to 1 3280/6561. The sum of the first 12 terms would be 265720/177147 or reduced to 1 88573/177147.
2006-11-02 05:37:21 · answer #3 · answered by Kevin B 2 · 0⤊ 0⤋
Formula is:
b^q - 1
---------
b-1
2006-11-02 05:28:45 · answer #4 · answered by Vladimir S 2 · 0⤊ 0⤋