Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. a) an = (an−1)2 , a1 = 2 ( Just find its general formula )

Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions. a) an = (an−1)2 , a1 = 2 ( Just find its general formula )

Question
Sequences
asked 2021-02-05
Find the first six terms of the sequence defined by each of these recurrence relations and initial conditions.
a) an = (an−1)2 , a1 = 2 ( Just find its general formula )

Answers (1)

2021-02-06
The given recurrence relation is: an=2an-1, a1=2
Note thet
a1=2
\(\displaystyle{a}{2}={2}{a}{1}={2}\cdot{2}={4}={2}^{{2}}\)
\(\displaystyle{a}{3}={2}{a}{2}={2}\cdot{4}={8}={2}^{{3}}\)
\(\displaystyle{a}{4}={2}{a}{3}={2}\cdot{8}={16}={2}^{{4}}\)
\(\displaystyle{a}{5}={2}{a}{4}={2}\cdot{16}={32}={2}^{{5}}\)
\(\displaystyle{a}{6}={2}{a}{5}={2}\cdot{32}={64}={2}^{{6}}\)
Therefore we find the general term is \(\displaystyle{a}{n}={2}^{{n}}\), for \(\displaystyle{n}∈{N}\). Prove: Note that
\(\displaystyle{a}{n}={2}{a}{n}-{1}={2}\cdot{2}\cdot{a}{n}-{2}={2}^{{2}}{a}{n}-{2}=\ldots={2}^{{n}}-{1}\cdot{a}{1}={2}^{{n}}\)
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