Calculus add formula to derive new formula

Question

I was asked to re-write a formula forward and backward and derive a new formula from it.

Here's the problem:

Write $$S=1+2+3+\cdots+N$$ forward and backward $$\begin{array}{rcrcrcr} S & = & 1 & + & 2 & +\cdots+ & N\\ S & = & N & + & N-1 & +\cdots+ & 1.\\ \overline{2S}&=&\overline{N+1}&+&\overline{N+1}&+\cdots+&\overline{N+1} \end{array}$$ Add vertically. Derive formula $(5.1.4)$ by solving for $S$.

$$\sum_{j=1}^N j=\dfrac{N(N+1)}2.\tag{5.1.4}$$

I'm not too sure where to start. Thanks!

Answer 1

So we have, as you see, after having summed the summation, $$2S = \underbrace{(N+1) + (N+1) +\cdots + (N+1)}_{N \text{ times}}$$

That is,

$$2S = N(N+1) \iff S = \frac{N(N+1)}{2}$$

Answer 2

There are $n$ terms, and each of them sum to $n+1$.

If we call our sum $S$, we know that $2S=n(n+1)$

From then on, it's a simple matter of rearranging to derive the formula that you need.