What is the sum of the arithmetic series 5 (sigma sign) n-1 (3n-2)?
It’s 35
Anonymous is correct. Remember, it's not cheating if you're not caught!
Pinetreestastelikemilk answer up until yes 1/2 are correct
7. 18
8. Yes; 4
9. 6144
10. +-6
11. 760
12. 21 n-1 (225-25n)
13. 35
14. 2
15. Converges; 12.5
the answers for the whole test shows up on the last question that I had to google wow
2,0,2,4
19
a1=-1an=an-1+5
an=6n-4
an=4n+4
-36
yes 1/2
6144
+-6
760
E(225-25n
35
2
convergese 12.5
felt
It's not ethical or honest to cheat on a test. It's important to learn and understand the material yourself. Cheating can have consequences such as getting a zero on the exam or even getting into trouble with your school. It's better to study and work hard to improve your own knowledge and understanding of the subject.
To find the sum of an arithmetic series, you can use the formula:
Sn = (n/2) * (a + l)
Where:
- Sn is the sum of the series,
- n is the number of terms,
- a is the first term, and
- l is the last term.
In the given arithmetic series, the first term (a) is 5, and the common difference is (3n - 2).
To find the last term (l), we substitute (n - 1) into the expression (3n - 2):
l = 3(n - 1) - 2
= 3n - 3 - 2
= 3n - 5
Now we can plug the values of a and l into the formula for Sn:
Sn = (n/2) * (a + l)
= (n/2) * (5 + 3n - 5)
= (n/2) * (3n)
= (3n^2)/2
Therefore, the sum of the arithmetic series 5 (sigma sign) (n - 1) (3n - 2) is given by (3n^2)/2.
Having n in the index and in the expression is bogus. If you mean
n-1
∑ (3k-2)
k=1
Then that can be split up into
∑3k - ∑2
Now, ∑2 is 2(n-1)
∑3k = 3∑k
You know that
n
∑k = n(n+1)/2
k=1
So for you example,
3∑k = 3(n-1)(n)/2
The final sum is thus
2(n-1) + 3(n-1)(n)/2
If I interpreted your notation wrong, then just fix it and use the above logic.