# Find a recurrence relation satisfied by this sequence. a_{n}=n^{2}

Find a recurrence relation satisfied by this sequence.
$$\displaystyle{a}_{{{n}}}={n}^{{{2}}}$$

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Clara Reese
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.