What is the sum of the geometric series
10
E 6(2)^n
n=1
I'm a little lost as to what I should do here. Could someone please clarify how this one is done?
1.D No
2.D 0.972
3.A 12 feet
4.C 250
5.A 10/sigma2n/n=1
6.D 10.A $138,037.53
7.A 1,365
8.A 19.662
9.C 12, 276
10.A $138,037.53
11.C It converges; it has a sum.
I made another one bc the first one was kinda confusing
Unit 4 Lesson 5 Geometric Series Alg 2 Connections
1.D No 6.D 165 11.C It converges; it has a sum.
2.D 0.972 7.A 1,365
3.A 12 feet 8.A 19.662
4.C 250 9.C 12, 276
5.A 10/sigma2n/n=1 10.A $138,037.53
You have a sequence with
a = 6*2^1 = 12
r=2
So, you have
S10 = 12 * (2^10-1)/(2-1) = 12276
All of the answers correct!!
^ same for honors except question 8 doesn't exist, so it's
8. C
9. A
10. C
thank you so much
THanks @KID as of 4/16/21
kid is still correct as of 2022
To find the sum of the given geometric series, you can use the formula for the sum of an infinite geometric series, which is given by:
S = a / (1 - r)
In this formula, 'S' represents the sum of the series, 'a' is the first term of the series, and 'r' is the common ratio between consecutive terms.
In the given series, we can see that the first term, 'a', is 10. The common ratio, 'r', can be determined by finding the ratio between consecutive terms:
r = 6(2)^n / 6(2)^(n-1)
Simplifying this expression:
r = 6(2)^n / (6(2)^(n-1))
= (2)^n / (2)^(n-1)
Applying the rule of exponents (when dividing with the same base, subtract the exponents), this can be further simplified:
r = 2^(n - (n - 1))
= 2^1
= 2
Now that we have the values of 'a' and 'r', we can substitute them into the formula to find the sum of the series:
S = 10 / (1 - 2)
= 10 / (-1)
= -10
Therefore, the sum of the geometric series is -10.