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

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.
