Find the summation of the following sequence Please give me an idea on how to proceed as I am really stuck with this, I have not encountered this type of question before yet my friend gave this summation to me and I am stuck.
$${4\choose 1} + \frac{5\choose2}{2}+\frac{6\choose3}{4}+\dots$$

Let $\displaystyle\binom{r+3}r\left(\dfrac12\right)^{r-1}=\binom{r+3}3\left(\dfrac12\right)^{r-1}=f(r+1)-f(r)$
where $f(m)=\left(\dfrac12\right)^m\sum_{r=0}^na_rm^r$ where $a_r$  are arbitrary constants
$\dfrac16(r+3)(r+2)(r+1)\left(\dfrac12\right)^{r-1}$
$=\left(\dfrac12\right)^{r+1}\left(a_0+a_1(r+1)+a_2(r+1)^2+a_3(r+1)^3+\cdots\right)-\left(\dfrac12\right)^r\left(a_0+a_1(r)+a_2(r)^2+a_3(r)^3+\cdots\right)$
$\dfrac{r^3+6r^2+11r+6}6$ $=\left(\dfrac12\right)^2\left(a_0+a_1(r+1)+a_2(r+1)^2+a_3(r+1)^3+\cdots\right)-\left(\dfrac12\right)\left(a_0+a_1(r)+a_2(r)^2+a_3(r)^3+\cdots\right)$
Clearly, $a_r=0$ for $r\ge4$
Comparing the coefficients of $r^3,$  $$\dfrac16=\dfrac{a_3}4-\dfrac{a_3}2\iff a_3=?$$
Comparing the coefficients of $r^2,$  $$1=\dfrac{a_2+3a_3}4-\dfrac{a_2}2\iff a_2=?$$
Similarly, comparing the coefficients of $r$ and the constants, we can find $a_1,a_0$
Using Telescoping Series, $$\sum_{r=1}^\infty\binom{r+3}3\left(\dfrac12\right)^{r-1}=-f(1)$$
