A simple recurrence problem: $L_n=L_{n-1}+n$ I am  studying concrete mathematics by Graham Knuth and Patashnik. In the first chapter lines in a plane he focuses on a equeation
$$L_n=L_{n-1}+n$$
on expanding
$$
\begin{align}
L_n&=L_{n-2}+(n-1)+n
\\&=L_{n-3}+(n-2)+(n-1)+n
\\&=\cdots
\\&=\cdots
\\&=L_0+1+2+3+\cdots+(n-1)+(n-2)
\end{align}
$$and so on.
Could someone explain how did the number $L_0+1+2+3+\cdots+(n-1)+(n-2)$ come into this series?
Thank you.
 A: It's just working out the equation starting with $L_n$ and ending up with $L_0$:
$$L_n = L_{n-1} + n = L_{n-2} + (n-1) + n = ... = L_2 + 3 + ... + (n-1) + n = L_1 + 2 + 3 + ... + (n-1) + n = L_0 + 1 + 2 + 3 + ... + (n-1) + n$$
Do you see it now?
A: From 
$$
{\rm L}_n-{\rm L}_{n-1}=n,\quad n=1,2,3,\cdots,
$$ one may just sum the following equalities:
$$
\begin{align}
{\rm L}_1-{\rm L}_0 &=1
\\{\rm L}_2-{\rm L}_1 &=2
\\{\rm L}_3-{\rm L}_2 &=3
\\{\cdots}\,{\cdots}
\\{\rm L}_n-{\rm L}_{n-1} &=n
\end{align}
$$ to get by a telescoping sum,
$$
{\rm L}_n-{\rm L}_{0}=1+2+3+\cdots+n.
$$
A: $$
L_n = L_{n-1} + n
$$
so we can write
$$
L_{n-1} = L_{(n-1)-1} + (n-1) = L_{n-2} + (n-1)
$$
we can put the first two lines together to form
$$
L_n = (L_{n-2} + (n-1)) + n
$$
we can keep this process going
$$
L_{n-2} = L_{n-3} + (n-2)
$$
or
$$
L_n = L_{n-3} + (n-2) + (n-1) + n = L_{n-3} + \sum_{k=0}^2 (n-k)
$$
or
$$
L_n = L_{n-10} + \sum_{k=0}^9 (n-k)
$$
or
$$
L_n = L_{n-m} + \sum_{k=0}^{m-1}(n-k) 
$$
lets take $m = n$ then we have
$$
L_n = \sum_{k=0}^{n-1}(n-k) 
$$
