How do I find the sum of a geometric series from 2 to infinity of -3/ (4^n) ?
a = -3/4
r = 1/4
S(2..∞) = S∞ - S1
so s=-13/4?
no it should be -1/4
To find the sum of an infinite geometric series, you can use the formula:
S = a / (1 - r)
Where:
- S is the sum of the series,
- a is the first term of the series,
- and r is the common ratio of the series.
In this case, the series is from 2 to infinity, so the first term (a) is -3/4^2 and the common ratio (r) is -3/4.
Now, substituting the values into the formula:
S = (-3/4^2) / (1 - (-3/4))
Simplifying the denominator:
S = (-3/4^2) / (1 + 3/4)
Simplifying the expression:
S = (-3/16) / (7/4)
Invert the denominator and multiply:
S = (-3/16) * (4/7)
S = -3/28
Therefore, the sum of the given geometric series from 2 to infinity is -3/28.