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

Oct 31, 2016

${\sum}_{i = 1}^{i = 5} \left(2 i + 1\right) = 35$
${\sum}_{i = 1}^{i = 5} \left(2 i + 1\right) = \left(2 \left(1\right) + 1\right) + \left(2 \left(2\right) + 1\right) + \left(2 \left(3\right) + 1\right) + \left(2 \left(4\right) + 1\right) + \left(2 \left(5\right) + 1\right)$
$\therefore {\sum}_{i = 1}^{i = 5} \left(2 i + 1\right) = \left(2 + 1\right) + \left(4 + 1\right) + \left(6 + 1\right) + \left(8 + 1\right) + \left(10 + 1\right)$
$\therefore {\sum}_{i = 1}^{i = 5} \left(2 i + 1\right) = 3 + 5 + 7 + 9 + 11$
$\therefore {\sum}_{i = 1}^{i = 5} \left(2 i + 1\right) = 35$