Let $P(n):(2 n+7) < (n+3)^2 \quad \ldots(i)$
For $n=1$, (i) becomes $2 \times 1+7 < (1+3)^2$ $\Rightarrow 9 < 16$, which is true
$\therefore P(1)$ is true
Let us suppose that $P(k)$ is true
i.e. $P(k):(2 k+7) < (k+3)^2 \quad \ldots(ii)$
Now, we will prove that $P(k+1)$ is true whenever $P(k)$ is true i.e.
$P(k+1): 2(k+1)+7 < (k+4)^2$
we know that $2 \mathrm{k}+7 < (\mathrm{k}+3)^2 \quad$ [from (ii)] $2 k+7+2 < (k+3)^2+2$
[Adding 2 on both sides]
$\begin{aligned}
&\Rightarrow \quad 2(k+1)+7 < (k+3)^2+2 \quad \ldots (iii)\\
&\Rightarrow \quad 2(k+1)+7 < k^2+6 k+11
\end{aligned}$
Adding $2 k+5$ to RHS
$\Rightarrow \quad 2(k+1)+7 < k^2+7 k+16$
or $2(k+1)+7 < (k+4)^2$
Thus $P(k+1)$ is true. Hence $P(n)$ is true for $n=k+1$
Hence, $P(n)$ is true for all natural numbers $n$.