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

Emily-Jane Bray 2021-09-16 Answered
Find a recurrence relation satisfied by this sequence.
\(\displaystyle{a}_{{{n}}}={n}^{{{2}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Clara Reese
Answered 2021-09-17 Author has 15770 answers
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.
Not exactly what you’re looking for?
Ask My Question
19
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...