How do you use sigma notation to write the sum for #[1-(1/6)^2]+[1-(2/6)^2]+...+[1-(6/6)^2]#?

Answer 1

#sum_(n=1)^6(1-(n/6)^2)#

If you look at the expression, the only thing that changes for each term is the number in red: #[1-(color(red)(1)/6)^2]+[1-(color(red)(2)/6)^2]+...+[1-(color(red)(6)/6)^2]#.

It starts from one and is incremented by one until it reaches 6. Everything else stays the same. We just need to substitute the number that is changing by a variable: #sum_(n=1)^6(1-(n/6)^2)#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Anctil

To express the given sum using sigma notation, we can first observe the pattern in the terms:

[1 - \left(\frac{1}{6}\right)^2 + 1 - \left(\frac{2}{6}\right)^2 + \ldots + 1 - \left(\frac{6}{6}\right)^2]

Notice that the expression (1 - \left(\frac{k}{6}\right)^2) appears in each term, where (k) ranges from 1 to 6. We can express this pattern using sigma notation.

The sum can be represented as:

[\sum_{k=1}^{6} \left(1 - \left(\frac{k}{6}\right)^2\right)]

This notation represents the sum of the expression (1 - \left(\frac{k}{6}\right)^2) as (k) ranges from 1 to 6.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.