Almost Consecutive

$The\quad sequence\quad is:\\ \Longrightarrow \sum _{ 0 }^{ \infty }{ \frac { (n+3)(n+2) }{ { 3 }^{ n }3! } } =\frac { 19 }{ 8 } \\ [break\quad the\quad terms\quad and\quad then\quad solve\quad it]$

$\begin{aligned} & & 1 + \frac {4}{6} + \frac {4 \cdot 5}{6\cdot 9}+ \frac {4 \cdot 5\cdot 6}{6\cdot 9\cdot 12} + \frac {4 \cdot 5\cdot 6 \cdot 7}{6\cdot 9\cdot 12 \cdot 15} + \ldots \\ & = & 3 \bigg [ \frac {1}{3} + \frac {4}{3 \cdot 6} + \frac {4 \cdot 5}{3 \cdot 6 \cdot 9} + \frac {4 \cdot 5 \cdot 6}{3 \cdot 6 \cdot 9 \cdot 12} + \ldots \bigg ] \\ & = & 3 \bigg [ \frac {3!/3!}{3^1 \cdot 1!} + \frac {4!/3!}{3^2 \cdot 2!} + \frac {5!/3!}{3^3 \cdot 3!} + \frac {6!/3!}{3^4 \cdot 4!} + \ldots \bigg ] \\ & = & 3 \sum_{k=1}^\infty \frac {(k+2)!/3!}{3^k \cdot k!} \\ & = & \frac {1}{2} \sum_{k=1}^\infty \frac {(k+2)(k+1)}{3^k} \\ \end{aligned}$

Which gives $\frac {19}{8}$ by using the properties of $\displaystyle \sum_{j=0}^\infty x^j = \frac {1}{1-x}$ for $\mid j \mid < 1$ .