Use Sigma Notation to write the sum
[10-(1/10)^2]+[10-(2/10)^2]+[10-(3/10)^2]+[10-(4/10)^2]
Just look at the pattern. The only thing that changes is the denominator.
would it not be ..
4
∑ (10 - i/10)^2
i=1
?
[10-(1/10)^2]+[10-(2/10)^2]+[10-(3/10)^2]+[10-(4/10)^2]
would it not be ..
4
∑ (10 - i/10)^2
i=1
?