Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 50

$16$

$a_{1} = 1$, $a_{n} = (5-n)a_{n-1}$ $a_{2} = (5-2)a_{1}$ $=3a_{1}$ $=3(1)$ $=3$ $a_{3} = (5-3)a_{2}$ $=(2)(3)$ $=6$ $a_{4}=(5-4)a_{3}$ $=6$ $a_{5}=(5-5)a_{4}=0$ $a_{6}=(5-6)a_{4}=0$ And all the other terms are zero Hence $\Sigma^{\infty}_{n=1}a_{n} = 1+ 3 +6 +6=16$