Find the sum:
8 over sigma symbol, k=1 below the symbol and (-1)^k+1,5k to the right of it.
Sorry the format is off, but I didnt know how else to type it in.
Thank you.
Here is how to type it.
The sum of (-1)^k +1.5k from k=1 to 8.
for eight terms, you can do this with your calc
k termvalue
1....0.5
2....4
3....3.5
4....7
5....6.5
6....10
7....9.5
8....13
See the pattern: each even add three to the previous even, and for the next odd, subtract .5
so predicting...
9....12.5 and you can check it.
Sum of the first 8 terms above, you can do that.
If you fish around, there are some generalized formulas for series such as this.
This is easy if you just break it up:
8
∑(-1)^k + 1.5k
1
= ∑(-1)^k + ∑1.5k
The first sum is -1+1-1+1... = 0, so
= ∑1.5k
= 1.5 ∑k
Now you know that
n
∑k = n(n+1)/2
1
so, your sum is just 1.5* 8*9/2 = 54
No problem! I understand what you're asking. The expression you provided is a summation notation. The sigma symbol (∑) represents the summation, and the subscript "k = 1" below the sigma symbol indicates that the variable "k" starts at 1. The expression (-1)^(k+1) is the alternating sign -1 raised to the power of (k+1), and 5k is the term being added in each iteration.
To find the sum, we can plug in the values of "k" and evaluate the expression for each value, then add all the terms together.
Let's break it down step by step:
1. Start with k = 1:
Evaluate (-1)^(k+1) = (-1)^(1+1) = (-1)^2 = 1
Evaluate 5k = 5 * 1 = 5
Add the values together: 1 * 5 = 5
2. Move to k = 2:
Evaluate (-1)^(k+1) = (-1)^(2+1) = (-1)^3 = -1
Evaluate 5k = 5 * 2 = 10
Add the values together: (-1) * 10 = -10
3. Continue this process by plugging different values for k, evaluating each term, and adding them together.
Let's calculate for a few more values of "k":
k = 3:
(-1)^(k+1) = (-1)^(3+1) = (-1)^4 = 1
5k = 5 * 3 = 15
(-1) * 15 = -15
k = 4:
(-1)^(k+1) = (-1)^(4+1) = (-1)^5 = -1
5k = 5 * 4 = 20
(-1) * 20 = -20
Now, add all the terms together:
5 + (-10) + (-15) + (-20) + ...
The pattern shows that the sign alternates between positive and negative with each iteration. To represent this pattern, we can see that (-1)^k+1 is equal to 1 when k is odd (1, 3, 5, ...) and -1 when k is even (2, 4, 6, ...).
Since the terms are alternating, this summation does not converge to a finite sum. It continues infinitely. However, we can still represent the partial sum. If you have a specific number of terms you want to sum, let me know, and I can help you calculate the partial sum.