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

1 Answer
Apr 6, 2017

Answer:

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

Explanation:

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)#.