Chapter 11 - Section 11.3 - Series - Exercise Set - Page 651: 33

$82$

Given that $ a_n=(n+3)(n+1) $ then the first 4 terms are \begin{array}{l} a_1=(1+3)(1+1)\\ a_1=(4)(2)\\ a_1=8 ,\\\\ a_2=(2+3)(2+1)\\ a_2=(5)(3)\\ a_2=15 ,\\\\ a_3=(3+3)(3+1)\\ a_3=(6)(4)\\ a_3=24 ,\\\\ a_4=(4+3)(4+1)\\ a_4=(7)(5)\\ a_4=35 .\end{array} Hence, the sum is $ 8+15+24+35 = 82 .$