Use the geometric sequence of numbers 1, 1/3, 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 terms? Carry all calculations to 6 decimals on all assignments.
Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms? Carry all calculations to 6 decimals on all assignments.
What observation can make about the successive partial sums of this series? In particular, what number does it appear that the sum will always be smaller than?
These are standard formulas. I will be happy to critique your thinking.
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S10 =a(r^n -1)/r-1
S10=? r=⅓÷1=⅓ a=1 n=10
By substitution,
S10=1(⅓^10-1)/⅓-1
S10 = -0.9999/-⅔
S10= 1.49985...
To find the sum of the first n terms of a geometric series, we can use the formula:
Sn = a(1 - r^n) / (1 - r)
where Sn is the sum of the first n terms, 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 equal to 1 and the common ratio r is equal to 1/3.
To find the sum of the first 10 terms, we substitute these values into the formula:
S10 = 1(1 - (1/3)^10) / (1 - 1/3)
Simplifying the expression inside the brackets and evaluating the final result to 6 decimal places, we get:
S10 = 1(1 - 0.000005) / (2/3) ≈ 0.999988
Therefore, the sum of the first 10 terms is approximately 0.999988.
To find the sum of the first 12 terms, we can use the same formula and substitute the values:
S12 = 1(1 - (1/3)^12) / (1 - 1/3)
Again, simplifying the expression inside the brackets and evaluating the final result to 6 decimal places, we get:
S12 = 1(1 - 0.000000015) / (2/3) ≈ 0.999999
Therefore, the sum of the first 12 terms is approximately 0.999999.
Observing the successive partial sums of this series, we notice that they seem to approach a certain number as more terms are added. In this case, it appears that the sum will always be smaller than 1.