A series involve combination I want find another Idea to find sum of $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ from $n=1 ,to,n=47$ 
or $$\sum_{n=1}^{47}\left(\begin{array}{c}n+3\\ 3\end{array}\right)=?$$ I do it first by turn $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ to $\dfrac{(n+3)(n+2)(n+1)}{3!}=\dfrac16 (n^3+6n^2+11n+6)$ and find sum of them by separation  $$\sum i=\dfrac{n(n+1)}{2}\\\sum i^2=\dfrac{n(n+1)(2n+1)}{6}\\\sum i^3=(\dfrac{n(n+1)}{2})^2$$ 
then I think more and do like below ... 
I think there is more Idea to find this summation . 
please hint, thanks in advanced
 A: By the well known hockey stick identity
$$ \sum_{n=0}^{47}\binom{n+3}{3} = \binom{47+3+1}{3+1} $$
and the problem is trivial from there.
A: This was my second try :
$$\sum_{n=1}^{47}\left(\begin{array}{c}n+3\\ 3\end{array}\right)=\\
\dfrac16\sum_{n=1}^{47}(n+3)(n+2)(n+1)=\\
\dfrac16.\dfrac14\sum_{n=1}^{47}(\color{red} {n+4-n})(n+3)(n+2)(n+1)=\\
\dfrac{1}{24}\sum_{n=1}^{47}(n+4)(n+3)(n+2)(n+1)-(n+3)(n+2)(n+1)n\\
\\by  \space {[f(n)=(n+3)(n+2)(n+1)n]}\\
\dfrac{1}{24}\sum_{n=1}^{47}f(n+1)-f(n)\\
\dfrac{1}{24}(f(48)-f(1))$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{n = 1}^{47}{n + 3 \choose  3} & =
-1 + \bracks{z^{47}}\sum_{k = 0}^{\infty}z^{k}
\bracks{\sum_{n = 0}^{k}{n + 3 \choose  n}} =
-1 + \bracks{z^{47}}
\sum_{n = 0}^{\infty}{-4 \choose  n}\pars{-1}^{n}\sum_{k = n}^{\infty}z^{k}
\\[5mm] & =
-1 + \bracks{z^{47}}
\sum_{n = 0}^{\infty}{-4 \choose  n}\pars{-1}^{n}{z^{n} \over 1 - z} =
-1 + \bracks{z^{47}}\bracks{{1 \over 1 - z}
\sum_{n = 0}^{\infty}{-4 \choose  n}\pars{-z}^{n}}
\\[5mm] & =
-1 + \bracks{z^{47}}\bracks{{1 \over 1 - z}\,\pars{1 - z}^{-4}} =
-1 + \bracks{z^{47}}\sum_{k = 0}^{\infty}{-5 \choose k}\pars{-z}^{k} =
-1 - {-5 \choose 47}
\\[5mm] & = \bbx{\ds{-1 + {51 \choose 47}}} =
-1 + {51 \times 50 \times 49 \times 48 \over 4 \times 3 \times 2} =
\bbx{\ds{249899}}
\end{align}
