Chapter 11 - Section 11.3 - Series - Exercise Set - Page 651: 38

$110$

Given that $ a_n=(n+4)^2 $ then the first 3 terms are \begin{array}{l} a_1=(1+4)^2\\ a_1=(5)^2\\ a_1=25 ,\\\\ a_2=(2+4)^2\\ a_2=(6)^2\\ a_2=36 ,\\\\ a_3=(3+4)^2\\ a_3=(7)^2\\ a_3=49 .\end{array} Hence, the sum is $ 25+36+49 = 110 .$