Recursion relation for the Laguerre polynomials How to come to the Laguerre recursion relation ,
$$(n+1)L_{n+1}^{(\alpha)}(x)+xL_n^{(\alpha)}(x)+ (n+\alpha) L_{n-1}^{(\alpha)}(x)=(2n+1+\alpha)L_n^{(\alpha)}(x) $$
from the sum for the generalized Laguerre polynomials :
$$(2n+1+\alpha) L_n^{(\alpha)} (x) = (2n+1+\alpha) \sum_{i=0}^n (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!}.$$
$$x L_n^{(\alpha)} (x) = - \sum_{i=1}^{n+1} (-1)^i {n+\alpha \choose n-i} \frac{x^{i}}{i!}\frac{i+\alpha}{n-i+1}i.$$
$$(n+\alpha) L_{n-1}^{(\alpha)} (x) =(n+\alpha) \sum_{i=0}^{n-1} (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!}\frac{n-i}{n+\alpha}.$$
$$(n+1) L_{n+1}^{(\alpha)} (x) = (n+1) \sum_{i=0}^{n+1} (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!}\frac{n+\alpha+1}{n-i+1}.$$
 A: The first term
\begin{eqnarray*}
(n+1) L_{n+1}^{(\alpha)} (x) &=&  \sum_{i=0}^{n+1} (-1)^i (\color{red}{n+1-i} +\color{blue}{i}){n+1+\alpha \choose n+1-i } \frac{x^i}{i!} \\
&=& \color{red}{(n+1+\alpha)} \sum_{i=0}^{n+1} (-1)^i {n+\alpha \choose n-i } \frac{x^i}{i!}  +\color{blue}{\sum_{i=0}^{n+1} (-1)^i {n+1+\alpha \choose n+1-i } \frac{x^i}{(i-1)!}}\\
&=& \color{red}{(n+1+\alpha)} \sum_{i=0}^{n+1} (-1)^i {n+\alpha \choose n-i } \frac{x^i}{i!} \\ &&+\color{blue}{\sum_{i=0}^{n+1} (-1)^i {n+\alpha \choose n+1-i } \frac{x^i}{(i-1)!}}+\color{green}{\sum_{i=0}^{n+1} (-1)^i {n+\alpha \choose n-i } \frac{x^i}{(i-1)!}}\\
\end{eqnarray*}
Note that the blue term above cancels with the second term in your formula.
Note that
\begin{eqnarray*}
(n+\alpha)  {n+\alpha-1 \choose n-i-1 } = (\color{red}{n}-\color{green}{i})  {n+\alpha \choose n-i } .
\end{eqnarray*}
Now the third term ...
\begin{eqnarray*}
(n+\alpha) L_{n-1}^{(\alpha)} (x) &=& \color{red}{n} \sum_{i=0}^{n-1} (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!} +\color{blue}{\frac{x^n}{(n-1)!}} \\ && - \color{blue}{\frac{x^n}{(n-1)!}} - \color{green}{ \sum_{i=0}^{n-1} (-1)^i {n+\alpha \choose n-i} \frac{x^i}{(i-1)!}} \\
&=& \color{red}{n} \sum_{i=0}^{\color{blue}{n}} (-1)^i {n+\alpha \choose n-i} \frac{x^i}{i!}  - \color{green}{ \sum_{i=0}^{\color{blue}{n}} (-1)^i {n+\alpha \choose n-i} \frac{x^i}{(i-1)!}}
\end{eqnarray*}
and we are pretty much done.
