A series involve combination

Question

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

Answer 1

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.

Answer 2

This was my second try : $$\sum_{n=1}^{47}\left(\begin{array}{c}n+3\\ 3\end{array}\right)=\\ \dfrac16\sum_{n=1}^{47}(n+3)(n+2)(n+1)=\\ \dfrac16.\dfrac14\sum_{n=1}^{47}(\color{red} {n+4-n})(n+3)(n+2)(n+1)=\\ \dfrac{1}{24}\sum_{n=1}^{47}(n+4)(n+3)(n+2)(n+1)-(n+3)(n+2)(n+1)n\\ \\by \space {[f(n)=(n+3)(n+2)(n+1)n]}\\ \dfrac{1}{24}\sum_{n=1}^{47}f(n+1)-f(n)\\ \dfrac{1}{24}(f(48)-f(1))$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}