Prove by mathematical induction
\[
3^n \gt n^2 \quad \text{for} \quad n \ge 2
\]
Prove the statement is true for a particular n. Here n = 2 is an obvious choice.
To show this is true for n=2, observe that the left hand side = 9 and the right hand side = 4.
So the statement is true for n=2.
Assume the statement is true for n = k and show it is true for n = k + 1.
To show the statement is true for n = k+1, show \( LHS - RHS \gt 0 \). So
\[ \begin{align}
LHS - RHS &= 3^{k+1} - (k+1)^2 \\
&= 3.3^k - (k^2 + 2k + 1) \quad (1)
\end{align} \ \]
Use the inductive hypothesis to simplify the right hand side of equation 1.
By the inductive hypothesis,
\( \quad 3.3^k - (k^2 + 2k + 1) \)
\( \quad \gt 3k^2 - (k^2 + 2k + 1) \)
\[ \begin{align}
&= 2k^2 - 2k - 1 \\
&= 2(k^2 - k - \frac{1}{2}) \\
&= 2[(k - \frac{1}{2})^2 - \frac{3}{4}] \quad \text{(by completing the square)}\\
&\gt 0 \quad \text{(for } k \ge 2)
\end{align} \ \]
So the statement is true for n = k + 1.
By the principle of mathematical induction,
The statement is true for \( n \ge 2 \).