Find the sum of the geometric series 128-64+32-______to 8 terms .
To find the sum of a geometric series, we can use the formula:
S = a * (1 - r^n) / ( 1 - r)
Where:
S is the sum of the series,
a is the first term,
r is the common ratio, and
n is the number of terms.
In this case, the first term (a) is 128, the common ratio (r) is -1/2 (since each term is divided by -2 to get the next term), and the number of terms (n) is 8.
Substituting these values into the formula, we get:
S = 128 * (1 - (-1/2)^8) / (1 - (-1/2))
Simplifying:
S = 128 * (1 - 1/256) / (3/2)
S = 128 * (255/256) * (2/3)
S = (128 * 255 * 2) / (256 * 3)
S = 65280 / 768
S = 85.
Therefore, the sum of the geometric series 128 - 64 + 32 - ... to 8 terms is 85.
To find the sum of a geometric series, we need to use the formula:
S = a * (r^n - 1) / (r - 1)
Where:
S is the sum of the series,
a is the first term,
r is the common ratio,
and n is the number of terms.
In this case, the first term (a) is 128, the common ratio (r) is -2 (since each term is multiplied by -2 to get the next term), and we are looking for the sum of 8 terms (n = 8).
Now, let me calculate it for you:
S = 128 * (-2^8 - 1) / (-2 - 1)
Simplifying this expression, we get:
S = 128 * (-256 - 1) / (-2 - 1)
S = 128 * (-257) / (-3)
S = -32896 / -3
S ≈ 10965.33
Therefore, the sum of the geometric series 128 - 64 + 32 - ______ (to 8 terms) is approximately 10965.33.