Evaluating $\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$ $$\frac{13}{1.2 .3 .2}+\frac{26}{2.3 .4 .4}+\frac{43}{3.4 .5 .8}+\frac{64}{4.5 .6 .16}+\cdots$$
I can reduce it to the general term,
$$\sum_{r=1}^\infty \frac{2r^2 + 7r +4}{r(r+1)(r+2)2^r}$$
I don't know how to go about this any further though. I also ran this in python and the sum is exceeding $1.5$ for $10,000$ terms, which is weird since it should converge to $1.5$, so it makes me doubt if the general term I've written is correct.

Using telescoping:
$$\begin{align}
\sum_{n=1}^\infty\frac{2n^2+7n+4}{n(n+1)(n+2)2^n}&=
\sum_{n=1}^\infty\left[\frac2n+\frac1{n+1}-\frac1{n+2}\right]\frac1{2^n}\\
&=\sum_{n=1}^\infty\left[\left(\frac1n+\frac1{n+1}\right)
-\frac12\left(\frac1{n+1}+\frac1{n+2}\right)\right]\frac1{2^{n-1}}\\
&=1+\frac12=\frac32.
\end{align}$$
