6 Evaluate the series 8Σn=3 5n. What is the lower limit of the summation natation is n=3
125
38
210
165
165
how did you get that?
To get the sum, we substitute the values of n into the expression 5n and add them up as follows:
8Σn=3 5n = 5(3) + 5(4) + 5(5) + ... + 5(10)
= 15 + 20 + 25 + ... + 50
We can see that this is an arithmetic series with first term a=15, common difference d=5, and last term l=50. We can use the formula for the sum of an arithmetic series:
Sn = (n/2)(a + l)
where n is the number of terms in the series. Since we are summing from n=3 to n=10, the number of terms is:
n = 10 - 3 + 1 = 8
Substituting the values, we get:
S8 = (8/2)(15 + 50) = 8(65) = 520
Therefore, the answer is 520 and the lower limit of the summation notation is n=3.
Show your work
6 Evaluate the series 8Σn=3 5n. What is the lower limit of the summation natation is n=3
125
38
210
165
9. What is the sum of the geometric series 10Σn=1 6(2)^n
756
12,276
6,138
The first term of the geometric series is:
a = 6(2)^1 = 12
The common ratio is:
r = 2
We can use the formula for the sum of a geometric series:
Sn = a(1 - r^n) / (1 - r)
Substituting the values, we get:
10Σn=1 6(2)^n = 6(2)^1 + 6(2)^2 + ... + 6(2)^10
= 12 + 24 + 48 + ... + 12,288
To find the number of terms in the series, we can use the formula:
l = a * r^(n-1)
where l is the last term of the series. Solving for n, we get:
n = log(l/a) / log(r) + 1
Substituting the values, we get:
n = log(12,288/12) / log(2) + 1 = 11
Therefore, the number of terms in the series is:
n = 11 - 1 + 1 = 11
Substituting again, we get:
Sn = 12(1 - 2^11) / (1 - 2) = 12(1 - 2048) / (-1) = 12(2047) = 24,564
Therefore, the sum of the geometric series is 24,564.
As usual, the bot messed up in the explanation
There are 6 terms, not 8, the last term is 40 , not 50
and then comes up with the wrong answer of course, even though
it did get the right answer of 165 for its first response.
For the first one , which the bot also messed up
you want Σ 5n , where n goes from 3 to 8
= 15 + 20 + 25 + 30 + 35 + 40 = (6/3)(15 + 40) = 165
for the second: Σ 6(2)^n , from 1 to 10
= 6(2^1) + 6(2^2) + .... + 6(2^10)
this is a GP with a = 12, r = 2 and we have 10 terms
sum(1) = a(r^10 - 1)/(2-1)
= 12(2^10 - 1)/1 = 12276