find the sum of the infinite geometric series
1/3^6+1/3^8+1/3^10+1/3^12........
-100/9+10/3-1+3/10
a = 1/3^6
r = 1/3^2
S = a/(1-r) = (1/3^6)/(1-1/9) = (1/3^6)(9/8) = 1/648
a = -100/9
r = -3/10
S = (-100/9)/(1+3/10) = -1000/117
thank you
To find the sum of an infinite geometric series, you need to use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where:
S represents the sum of the series,
a represents the first term in the series, and
r represents the common ratio between consecutive terms.
In the case of the first series:
a = 1/3^6
r = 1/3^2 = 1/9
Plugging these values into the formula for S:
S = (1/3^6) / (1 - 1/9)
To simplify this expression, let's first find a common denominator for the denominator:
S = (1/3^6) / (9/9 - 1/9)
S = (1/3^6) / (8/9)
Next, we invert the denominator and multiply:
S = (1/3^6) * (9/8)
S = 9/((3^6) * 8)
By simplifying the expression further, we get:
S = 9 / (729 * 8)
S = 9/5832
Therefore, the sum of the infinite geometric series 1/3^6 + 1/3^8 + 1/3^10 + 1/3^12 + ... is equal to 9/5832.
Now let's move on to the second series:
-100/9 + 10/3 - 1 + 3/10
To find the sum directly, you can simply add the terms together:
(-100/9) + (10/3) - 1 + (3/10) = -100/9 + 30/9 - 9/9 + 27/90
Combining the terms with a common denominator:
= (-100 + 30 - 9 + 27)/9
= -52/9
Therefore, the sum of the series -100/9 + 10/3 - 1 + 3/10 is equal to -52/9.