Write the sum using sigma notation.
-3/8 + 4/7 - 5/6 + 6/5 - 7/4+ 8/3
Rob/Jeff -- please use the same name for your posts.
I assume you know how to write sigma: ∑
Note that the numerators go from 3 to 8, alternating signs, and the denominators count down from 8 to 3:
8
∑ (-1)^k * k/(11-k)
k=3
To write the sum using sigma notation, we need to find a pattern in the given terms.
Looking at the terms, we can observe that the numerator of each term is a positive integer that increases by 1 for each subsequent term. The denominator is also a positive integer that starts from 8 and decreases by 1 for each subsequent term.
Let's break this down step by step:
First, let's establish the pattern for the numerator: starting with -3, each subsequent term increases by 1. We can represent this with the variable 'n', where n is the position of the term.
Second, let's establish the pattern for the denominator: starting with 8, each subsequent term decreases by 1. We can also represent this with the variable 'n'.
Now we can write the sigma notation:
The sum can be written as:
∑ (n=1 to 6) [(n+2)/(n+5)]
Here, the value of 'n' ranges from 1 to 6, representing the position of each term. (n=1 for the first term, n=2 for the second term, and so on, up to n=6 for the sixth term).
The expression inside the summation, [(n+2)/(n+5)], represents each term in the series, where 'n' takes on the values 1 to 6.
So, the sum can be simplified as:
(-1/8) + (6/7) + (7/6) + (8/5) + (9/4) + (10/3)
Note that we added 2 to each numerator and 5 to each denominator to account for the pattern we observed.
Thus, the sum of the series from the given terms, written using sigma notation, is ∑ (n=1 to 6) [(n+2)/(n+5)].