<s>[INST] How is this expression derived? On Concrete Mathematics (page 7) we have the following:
$L_n = L_{n-1} + n = (\frac{1}{2}(n-1)n + 1 ) + n = \frac{1}{2}n(n+1)+1$
How is the last expression derived? 
How do you convert $(\frac{1}{2}(n-1)n + 1 ) + n$ into $\frac{1}{2}n(n+1)+1$ ?
I've tried different things but was not able to solve it.
 [/INST] Multiply out, collect the first-degree terms, and then put $\frac12n$ back outside a parenthesis again:
$$ \begin{align} (\tfrac12 (n-1)n + 1) + n &=
\tfrac12 n^2 - \tfrac12 n + 1 + n \\&=
\tfrac12 n^2 + n - \tfrac12n + 1 \\&=
\tfrac12 n^2 + \tfrac12 n + 1 \\&=
\tfrac12n (n + 1) + 1
\end{align} $$
Or you could also do it as:
$$ \begin{align} (\tfrac12 (n-1)n + 1) + n &=
\tfrac12 n(n-1) + 1  + \tfrac 12n \cdot 2 \\&=
\tfrac12n(n-1+2) + 1 \\&=
\tfrac12n (n + 1) + 1
\end{align} $$
</s>