Question

How do you find the sum given `Sigma(2i+1)` from i=1 to 5?

Answer 1

`sum_(i=1)^(i=5)(2i+1) = 35`

Explanation:

`sum_(i=1)^(i=5)(2i+1) = (2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1) + (2(5)+1)`
`:. sum_(i=1)^(i=5)(2i+1) = (2+1) + (4+1) + (6+1) + (8+1) + (10+1)`
`:. sum_(i=1)^(i=5)(2i+1) = 3 + 5 + 7 + 9 + 11`
`:. sum_(i=1)^(i=5)(2i+1) = 35`