find the sum of each of the infinite series.
A. 9 - 3 + 1 - (1/3) + (1/9) ...
B. (1/2) + 1 +2 + 4 ....
2007-07-14 12:19:20 · 4 answers · asked by Sara 3 in Science & Mathematics ➔ Mathematics
9 - 3 + 1 - (1/3) + (1/9) ...
This is a alternating geometric series with R = - 1/3
Hence it's sum coverges to a value 6+1/(1-R) = 6+ 1/(1+1/3) = 6 3/4
B. (1/2) + 1 +2 + 4 ....
Here the geometric series has r=2. Thus the series diverges and has no finite sum. |R| must be <1.
2007-07-14 12:31:41 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
sum = 9 /(1 - 1/9) - 3 / (1 - 1/9)
. . . . = 9 / (8/9) - 3 / (8/9)
. . . . = 6 (9/8) = 6.75
Sum = sum of 2 n . . . starting from 1/2
2007-07-14 12:30:44 · answer #2 · answered by CPUcate 6 · 0⤊ 1⤋
These series are Geometric progressions.
So use the formula Sinf = a/(1-r) if abs(r)<1
I think!!
A. a=9, r=-1/3
Sinf = 9/(1-(-1/3))
Sinf= 9/(4/3) = 27/4
b. a = 1/2, r=2
hey this one will keep on getting bigger since r>1
So sinf = inf
2007-07-14 12:28:10 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(A)
Sum = 9/(1-(-1/3))
=6.75
(B)
Sum = infinity
2007-07-14 12:25:48 · answer #4 · answered by A 150 Days Of Flood 4 · 0⤊ 0⤋