Asked by slayrat
use summation notation to write each arithmetic series for the specifies number of terms. Then evaluate the sum. 10+7+4+...; n=5
Answers
Answered by
Bot
The arithmetic series can be written as:
$\sum_{i=1}^{5} (10-3(i-1))$
To evaluate the sum, we can simplify the expression inside the summation:
$10-3(i-1) = 10-3i+3$
$10-3(i-1) = 13-3i$
Now we can rewrite the sum using this simplified expression:
$\sum_{i=1}^{5} (13-3i)$
$= (13-3\cdot1) + (13-3\cdot2) + (13-3\cdot3) + (13-3\cdot4) + (13-3\cdot5)$
$= 13-3+13-6+13-9+13-12+13-15$
$= 60$
Therefore, the sum of the arithmetic series 10+7+4+... (with n=5) is 60.
$\sum_{i=1}^{5} (10-3(i-1))$
To evaluate the sum, we can simplify the expression inside the summation:
$10-3(i-1) = 10-3i+3$
$10-3(i-1) = 13-3i$
Now we can rewrite the sum using this simplified expression:
$\sum_{i=1}^{5} (13-3i)$
$= (13-3\cdot1) + (13-3\cdot2) + (13-3\cdot3) + (13-3\cdot4) + (13-3\cdot5)$
$= 13-3+13-6+13-9+13-12+13-15$
$= 60$
Therefore, the sum of the arithmetic series 10+7+4+... (with n=5) is 60.
Answered by
Bot
Yes, that is correct.
Answered by
Bot
Yes, I'm sure. The solution I provided uses the summation notation to write the arithmetic series and then simplifies it to evaluate the sum. The final answer is 60.
Answered by
Bot
I apologize for the mistake in my earlier response. You are correct that the sum of the arithmetic series 10+7+4+1+(-2) (with n=5) is 20. The series can be written using summation notation as:
$\sum_{i=1}^{5} (10-3(i-1))$
Simplifying the expression inside the summation, we get:
$10-3(i-1) = 13-3i$
So the sum can be written as:
$\sum_{i=1}^{5} (13-3i)$
$= (13-3\cdot1) + (13-3\cdot2) + (13-3\cdot3) + (13-3\cdot4) + (13-3\cdot5)$
$= 13-3+13-6+13-9+13-12+13-15$
$= 20$
I hope this clears up any confusion. Thank you for bringing this to my attention.
$\sum_{i=1}^{5} (10-3(i-1))$
Simplifying the expression inside the summation, we get:
$10-3(i-1) = 13-3i$
So the sum can be written as:
$\sum_{i=1}^{5} (13-3i)$
$= (13-3\cdot1) + (13-3\cdot2) + (13-3\cdot3) + (13-3\cdot4) + (13-3\cdot5)$
$= 13-3+13-6+13-9+13-12+13-15$
$= 20$
I hope this clears up any confusion. Thank you for bringing this to my attention.
Answered by
Bot
Yes, that is another way to evaluate the sum of the arithmetic series. We can use the formula for the sum of an arithmetic series, which is:
$S_n = \frac{n}{2}(a_1 + a_n)$
where $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the nth term.
For the given series 10+7+4+1+(-2), we have $n=5$, $a_1=10$, and $a_n=-2$. So the formula becomes:
$S_5 = \frac{5}{2}(10 + (-2))$
$= \frac{5}{2}(8)$
$= 20$
This matches the previous result we obtained using the summation notation. Thank you for providing this alternative solution.
$S_n = \frac{n}{2}(a_1 + a_n)$
where $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the nth term.
For the given series 10+7+4+1+(-2), we have $n=5$, $a_1=10$, and $a_n=-2$. So the formula becomes:
$S_5 = \frac{5}{2}(10 + (-2))$
$= \frac{5}{2}(8)$
$= 20$
This matches the previous result we obtained using the summation notation. Thank you for providing this alternative solution.
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