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

floymdiT 2021-09-19 Answered
Find a recurrence relation satisfied by this sequence.
\(\displaystyle{a}_{{{n}}}={n}^{{{2}}}+{n}\)

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Expert Answer

Layton
Answered 2021-09-20 Author has 17662 answers
Given:
\(\displaystyle{a}_{{{n}}}={n}^{{{2}}}+{n}\)
Let us first determine the first term by replacing n in the given expression for \(\displaystyle{a}_{{{n}}}\) by 0:
\(\displaystyle{a}_{{{0}}}={0}^{{{2}}}+{0}={0}\)
Let us similarly determine the next few terms as well:
\(\displaystyle{a}_{{{1}}}={1}^{{{2}}}+{1}={2}={a}_{{{0}}}+{2}{\left({1}\right)}\)
\(\displaystyle{a}_{{{2}}}={2}^{{{2}}}+{2}={6}={a}_{{{1}}}+{2}{\left({2}\right)}\)
\(\displaystyle{a}_{{{3}}}={3}^{{{2}}}+{3}={12}={a}_{{{2}}}+{2}{\left({3}\right)}\)
\(\displaystyle{a}_{{{4}}}={4}^{{{2}}}+{4}={20}={a}_{{{3}}}+{2}{\left({4}\right)}\)
\(\displaystyle{a}_{{{5}}}={5}^{{{2}}}+{5}={30}={a}_{{{4}}}+{2}{\left({5}\right)}\)
\(\displaystyle{a}_{{{6}}}={6}^{{{2}}}+{6}={42}={a}_{{{5}}}+{2}{\left({6}\right)}\)
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.
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