Write out this sum twice, once is direct order, and once in reverse:

\(\displaystyle{1}+{2}+\ldots+{\left({n}-{1}\right)}+{n}={s}\)

\(\displaystyle{n}+{\left({n}-{1}\right)}+\ldots+{2}+{1}={s}\)

Now add up column-wise:

\(\displaystyle{\left({n}+{1}\right)}+{\left({n}+{1}\right)}+\ldots+{\left({n}+{1}\right)}+{\left({n}+{1}\right)}={2}{s}\)

There are exactly n terms here (as many as the number of terms in the sum). Hence:

\(\displaystyle{n}{\left({n}+{1}\right)}={2}{s}\)

Now solve for s.

\(\displaystyle{1}+{2}+\ldots+{\left({n}-{1}\right)}+{n}={s}\)

\(\displaystyle{n}+{\left({n}-{1}\right)}+\ldots+{2}+{1}={s}\)

Now add up column-wise:

\(\displaystyle{\left({n}+{1}\right)}+{\left({n}+{1}\right)}+\ldots+{\left({n}+{1}\right)}+{\left({n}+{1}\right)}={2}{s}\)

There are exactly n terms here (as many as the number of terms in the sum). Hence:

\(\displaystyle{n}{\left({n}+{1}\right)}={2}{s}\)

Now solve for s.