Given:

\(\displaystyle{a}_{{{n}}}={n}^{{{2}}}\)

Let us first determine the first term by replacing n in the given expression for \(\displaystyle{a}_{{{n}}}\) by 0:

\(\displaystyle{a}_{{{0}}}={0}^{{{2}}}={0}\)

Let us similarly determine the next few terms as well:

\(\displaystyle{a}_{{{1}}}={1}^{{{2}}}={1}={a}_{{{0}}}+{2}{\left({1}\right)}-{1}\)

\(\displaystyle{a}_{{{2}}}={2}^{{{2}}}={4}={a}_{{{1}}}+{2}{\left({2}\right)}-{1}\)

\(\displaystyle{a}_{{{3}}}={3}^{{{2}}}={9}={a}_{{{2}}}+{2}{\left({3}\right)}-{1}\)

\(\displaystyle{a}_{{{4}}}={4}^{{{2}}}={16}={a}_{{{3}}}+{2}{\left({4}\right)}-{1}\)

\(\displaystyle{a}_{{{5}}}={5}^{{{2}}}={25}={a}_{{{4}}}+{2}{\left({5}\right)}-{1}\)

\(\displaystyle{a}_{{{6}}}={6}^{{{2}}}={36}={a}_{{{5}}}+{2}{\left({6}\right)}-{1}\)

We note that each term is the previous term increased by 2n-1:

\(\displaystyle{a}_{{{n}}}={a}_{{{n}-{1}}}+{2}{n}-{1}\)

Thus a recurrence relation for \(\displaystyle{a}_{{{n}}}\) is then:

\(\displaystyle{a}_{{{0}}}={0}\)

\(\displaystyle{a}_{{{n}}}={a}_{{{n}-{1}}}+{2}{n}-{1}\)

Note: There are infinitely many different recurence relations that satisfy any sequence.

\(\displaystyle{a}_{{{n}}}={n}^{{{2}}}\)

Let us first determine the first term by replacing n in the given expression for \(\displaystyle{a}_{{{n}}}\) by 0:

\(\displaystyle{a}_{{{0}}}={0}^{{{2}}}={0}\)

Let us similarly determine the next few terms as well:

\(\displaystyle{a}_{{{1}}}={1}^{{{2}}}={1}={a}_{{{0}}}+{2}{\left({1}\right)}-{1}\)

\(\displaystyle{a}_{{{2}}}={2}^{{{2}}}={4}={a}_{{{1}}}+{2}{\left({2}\right)}-{1}\)

\(\displaystyle{a}_{{{3}}}={3}^{{{2}}}={9}={a}_{{{2}}}+{2}{\left({3}\right)}-{1}\)

\(\displaystyle{a}_{{{4}}}={4}^{{{2}}}={16}={a}_{{{3}}}+{2}{\left({4}\right)}-{1}\)

\(\displaystyle{a}_{{{5}}}={5}^{{{2}}}={25}={a}_{{{4}}}+{2}{\left({5}\right)}-{1}\)

\(\displaystyle{a}_{{{6}}}={6}^{{{2}}}={36}={a}_{{{5}}}+{2}{\left({6}\right)}-{1}\)

We note that each term is the previous term increased by 2n-1:

\(\displaystyle{a}_{{{n}}}={a}_{{{n}-{1}}}+{2}{n}-{1}\)

Thus a recurrence relation for \(\displaystyle{a}_{{{n}}}\) is then:

\(\displaystyle{a}_{{{0}}}={0}\)

\(\displaystyle{a}_{{{n}}}={a}_{{{n}-{1}}}+{2}{n}-{1}\)

Note: There are infinitely many different recurence relations that satisfy any sequence.