<s>[INST] Solving $nx_n=(n+2)x_{n-1} + 1$ by the telescoping method I am trying to solve this recurrence relation from a book "Problem solving through Problems" by Loren c. Larson (5.3.14 (b)) using the telescoping method.
$$x_0=0\qquad nx_n=(n+2)x_{n-1} + 1\ (n > 0)$$
Here some additional terms I computed: $x_1=1, x_2 = 5/2, x_3 = 9/2, x_4=7$.
So far I tried to sum both parts of the expression for all $x_i$ and indeed, after some telescoping I got this:
$$x_n = \frac3n(x_{n-1} + \ldots + x_1) + 1$$
Although it seems of no help to actually get the result.
 [/INST] Dividing both parts of the expression by $n(n+1)(n+2)$, one gets:
$$\frac{x_n}{(n+2)(n+1)}=\frac{x_{n-1}}{(n+1)n} + \frac{1}{(n+2)(n+1)n}$$
Let us define $$y_n = \frac{x_n}{(n+2)(n+1)}$$
Then the new recurrence is easier to solve:
$$y_n = y_{n-1} + \frac{1}{(n+2)(n+1)n}\qquad y_0 = 0$$
One knows that there exists $(A,B,C)$ such that $$\frac{1}{(n+2)(n+1)n}=\frac{A}{n+2} + \frac{B}{n+1} + \frac{C}{n}$$
One easily checks that $A = 1/2$, $B = -1$, $C = 1/2$, hence
 our recurrence takes the following form:
$$y_n = y_{n-1} + \frac{1/2}{n+2} - \frac{1}{n+1} + \frac{1/2}{n}$$
It follows that
$$y_1 - y_0 = \color{red}{\frac{1/2}{3}} - \frac{1}{2} + \frac{1/2}{1}$$
$$y_2 - y_1 = \color{blue}{\frac{1/2}{4}} - \color{red}{\frac{1}{3}} + \frac{1/2}{2}$$
$$y_3 - y_2 = \color{green}{\frac{1/2}{5}} - \color{blue}{\frac{1}{4}} + \color{red}{\frac{1/2}{3}}$$
$$y_4 - y_3 = \color{orange}{\frac{1/2}{6}} - \color{green}{\frac{1}{5}} + \color{blue}{\frac{1/2}{4}}$$
$$\ldots$$
$$y_{n-2} - y_{n-3} = \color{orange}{\frac{1/2}{n}} - \color{#08f}{\frac{1}{n-1}} + \color{#08A}{\frac{1/2}{n-2}}$$
$$y_{n-1} - y_{n-2} = \frac{1/2}{n+1} - \color{orange}{\frac{1}{n}} + \color{#08f}{\frac{1/2}{n-1}}$$
$$y_{n} - y_{n-1} = \frac{1/2}{n+2} - \frac{1}{n+1} + \color{orange}{\frac{1/2}{n}}$$
Now if we sum both parts of all the expressions notice how terms of the same color cancel each other. Also all $y_i$ for the exception of $y_0$ and $y_n$ will be canceled as well.
Eventually we get 
$$y_n = \frac14 - \frac{1}{2(n+1)} + \frac{1}{2(n+2)}$$
Then going back to $x_n$ we get
$$x_n = \frac{n^2 + 3n}{4}$$
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