Prove the following by using the principle of mathematical induction for all $n \in \mathrm{N}$ :
$(2 n+7)<(n+3)^{2}$
Let the given statement be P(n), i.e.,
$\mathrm{P}(n):(2 n+7)<(n+3)^{2}$
It can be observed that $P(n)$ is true for $n=1$ since $2.1+7=9<(1+3)^{2}=16$, which is true.
Let P(k) be true for some positive integer k, i.e.,
$(2 k+7)<(k+3)^{2}$ (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider
$\{2(k+1)+7\}=(2 k+7)+2$
$\therefore\{2(k+1)+7\}=(2 k+7)+2<(k+3)^{2}+2$ [using (1)]
$2(k+1)+7
$2(k+1)+7
Now, $k^{2}+6 k+11
$\therefore 2(k+1)+7<(k+4)^{2}$
$2(k+1)+7<\{(k+1)+3\}^{2}$
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.