question about Laguerre polynomials how to prove that 
$$L_{n+1}(x)=\frac{1}{n+1}((2n+1-x)L_n(x)-nL_{n-1}(x))$$
I see it on wikipedia but I dont know how they prove it
 A: Here's an elegant proof, using specific characteristics of Laguerre polynomials.
Rodrigues' formula is
$$L_n(x) = \frac{e^x}{n!}[e^{-x}x^n]^{(n)}$$
$L_n$ is a polynomial of degree $n$, and writing $L_n(x) = \sum_{j=0}^{n} a_jx^j$, it is easy to derive
\begin{cases}
a_0 = 1
\\a_1 = -n
\\a_n = (-1)^n/n!
\end{cases}
$L_n$ and $L_m$ are orthogonal when $l \not= m$, meaning that their internal product is zero, when using the definition $(f,g) = \int_0^\infty e^{-x}fgdx$. This internal product has the special characteristic that $(xf, g) = (f, xg)$.
Now $xL_n$ is a polynomial of degree $n+1$, and thus it is a linear combination of $L_{n+1}, L_n, ..., L_1$.
When $m+1 < n$, then, because $xL_m$ is a linear combination of $L_{m+1}, L_m, ..., L_1$, we derive that $(L_n, xL_m) = 0$, hence also $(xL_n, L_m) = 0$.
We conclude that $xL_n$ is a linear combination of $L_{n+1}, L_n, L_{n-1}$ only.
Writing $$xL_n = \alpha L_{n+1} + \beta L_n + \gamma L_{n-1}$$ and looking at the coefficients of the terms 1, $x$ and $x^{n+1}$, we see that
\begin{cases}
0 = \alpha + \beta +\gamma
\\1 = \alpha\cdot -(n+1) + \beta\cdot -n + \gamma\cdot -(n-1) 
\\(-1)^n/n! = (-1)^{n+1}\alpha/(n+1)!
\end{cases}
so
\begin{cases}
\alpha = -(n+1)
\\0 = \alpha + \beta +\gamma
\\1 = -n(\alpha + \beta +\gamma) - \alpha + \gamma = -\alpha + \gamma
\end{cases}
so
\begin{cases}
\alpha = -(n+1)
\\\gamma = 1 + \alpha = -n
\\\beta = -(\alpha + \gamma) = 2n + 1
\end{cases}
and we conclude
$$xL_n = -(n+1)L_{n+1} + (2n + 1)L_n - nL_{n-1}$$
or
$$(n+1)L_{n+1} = (2n + 1 - x)L_n - nL_{n-1}$$
