How is this expression derived?

Question

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.

Answer 1

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: $$ \begin{align} (\tfrac12 (n-1)n + 1) + n &= \tfrac12 n(n-1) + 1 + \tfrac 12n \cdot 2 \\&= \tfrac12n(n-1+2) + 1 \\&= \tfrac12n (n + 1) + 1 \end{align} $$

Answer 2

We say about the Jakob Steiner's problem.

There written: $$L_n=L_{n-1}+n=L_{n-2}+(n-1)+n=...=L_0+1+2+...+(n-1)+n=1+\frac{n(n+1)}{2}.$$ Because $L_0=1$ and $1+2+...+n=\frac{n(n+1)}{2}.$