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

\(\displaystyle{a}{n}={2}{a}{n}-{1}={2}\cdot{2}\cdot{a}{n}-{2}={2}^{{2}}{a}{n}-{2}=\ldots={2}^{{n}}-{1}\cdot{a}{1}={2}^{{n}}\)

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

\(\displaystyle{a}{n}={2}{a}{n}-{1}={2}\cdot{2}\cdot{a}{n}-{2}={2}^{{2}}{a}{n}-{2}=\ldots={2}^{{n}}-{1}\cdot{a}{1}={2}^{{n}}\)