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.
$${4\choose 1} + \frac{5\choose2}{2}+\frac{6\choose3}{4}+\dots$$
 A: Oh.
I misread the denominator as $n+1$,
not $2^n$.
So this is easy.
Note that $\binom{n}{n-3}  =\binom{n}{3}$.
Then look at $(1-x)^{-a}$
for $a=4$. 

Here's more.
Your series is
$\sum_{n=1}^{\infty} \frac1{2^{n-1}}\binom{n+3}{n}
=2\sum_{n=1}^{\infty} \frac1{2^n}\binom{n+3}{3}
$.
Consider
$f_a(x)
=(1-x)^{-a}
=\sum_{n=0}^{\infty} \binom{a+n-1}{n} x^n
$.
(see https://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem)
Your series is
$\begin{array}\\
2\sum_{n=1}^{\infty} \frac1{2^n}\binom{n+3}{n}
&=2(-1+\sum_{n=0}^{\infty} \frac1{2^n}\binom{n+3}{n})\\
&=2(f_4(\frac12)-1)\\
&=2(\dfrac1{(1-\frac12)^4}-1)\\
&=2(\dfrac1{\frac12^4}-1)\\
&=30\\
\end{array}
$
A: $$\sum_{r=1}^\infty\binom{r+3}r\left(\dfrac12\right)^{r-1}=\sum_{r=1}^\infty\binom{r+3}3\left(\dfrac12\right)^{r-1}$$
$$=\dfrac26\sum_{r=1}^\infty\left(r^3\left(\dfrac12\right)^r+6r^2\left(\dfrac12\right)^r+11r\left(\dfrac12\right)^r+6\left(\dfrac12\right)^r\right)$$
Now  for $|x|<1,$  $$\sum_{r=0}^\infty ax^r=\dfrac a{1-x}$$
We need repeated differentiation 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{n = 1}^{\infty}{{n + 3 \choose n} \over 2^{n - 1}} & =
-2 + 2\sum_{n = 0}^{\infty}{1 \over 2^{n}}\
\overbrace{{-\bracks{n + 3} + n - 1 \choose n}\pars{-1}^{n}}^{\ds{\mbox{Binomial Negation}}}\ =\
-2 + 2\sum_{n = 0}^{\infty}
{-4 \choose n}\pars{-\,{1 \over 2}}^{n}
\\[5mm] & =
-2 + 2\bracks{1 + \pars{-\,{1 \over 2}}}^{-4} = \bbx{\ds{30}}
\end{align}
A: 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, $$\sum_{r=1}^\infty\binom{r+3}3\left(\dfrac12\right)^{r-1}=-f(1)$$
