Claim: \(\displaystyle{3}^{{{2}{n}}}-{1}\) is divisible by 4 for \(\displaystyle{n}\in{N}\).

We will prove this by induction.

Base step: n = 1

\(\displaystyle\therefore{3}^{{2}}-{1}={8}\), which is divisible by 4

\(\displaystyle\therefore\)the result is true fir n =1.

Inductive step:

Assure that result is true for n = k

i.e. \(\displaystyle\frac{{4}}{{3}^{{{2}{k}}}}-{1}\)...(1)

Now, we prove that: result is true form = k+1

i.e. \(\displaystyle\frac{{4}}{{3}^{{{2}{\left({k}+{1}\right)}}}}-{1}\)

Consider \(\displaystyle{3}^{{{2}{\left({k}+{1}\right)}}}\equiv{3}^{{{2}{k}}}\cdot{9}{\left(\text{mod}{4}\right)}\)

\(\displaystyle\equiv{1}\cdot{9}{\left(\text{mod}{4}\right)}{\left[{\mathfrak{{o}}}{m}{\left({i}\right)}\right]}\)

\(\displaystyle\therefore{3}^{{{2}{\left({k}+{1}\right)}}}\equiv{1}{\left(\text{mod}{4}\right)}\)

\(\displaystyle\therefore\frac{{4}}{{3}^{{{2}{\left({k}+{1}\right)}}}}-{1}\)

\(\displaystyle\therefore\)The result is true for n = k+1.

\(\displaystyle\therefore\)By induction,

\(\displaystyle{3}^{{{2}{\left({n}+{1}\right)}}}-{1}\) is divisibly by 4 for \(\displaystyle{n}\in{N}\)

We will prove this by induction.

Base step: n = 1

\(\displaystyle\therefore{3}^{{2}}-{1}={8}\), which is divisible by 4

\(\displaystyle\therefore\)the result is true fir n =1.

Inductive step:

Assure that result is true for n = k

i.e. \(\displaystyle\frac{{4}}{{3}^{{{2}{k}}}}-{1}\)...(1)

Now, we prove that: result is true form = k+1

i.e. \(\displaystyle\frac{{4}}{{3}^{{{2}{\left({k}+{1}\right)}}}}-{1}\)

Consider \(\displaystyle{3}^{{{2}{\left({k}+{1}\right)}}}\equiv{3}^{{{2}{k}}}\cdot{9}{\left(\text{mod}{4}\right)}\)

\(\displaystyle\equiv{1}\cdot{9}{\left(\text{mod}{4}\right)}{\left[{\mathfrak{{o}}}{m}{\left({i}\right)}\right]}\)

\(\displaystyle\therefore{3}^{{{2}{\left({k}+{1}\right)}}}\equiv{1}{\left(\text{mod}{4}\right)}\)

\(\displaystyle\therefore\frac{{4}}{{3}^{{{2}{\left({k}+{1}\right)}}}}-{1}\)

\(\displaystyle\therefore\)The result is true for n = k+1.

\(\displaystyle\therefore\)By induction,

\(\displaystyle{3}^{{{2}{\left({n}+{1}\right)}}}-{1}\) is divisibly by 4 for \(\displaystyle{n}\in{N}\)