# Prove the following by using the principle of mathematical induction for all $n \in N$  $(2n+7) < (n+3)^2$

Let the given statement be $P(n)$, i.e.,
$P(n) : (2n+7) < (n+3)^2$
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.,
$(2k+7)<(k+3)^2$----------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Consider
$\{2(k+1)+7\} = (2k+7)+2$
$\therefore \{2(k+1)+7\} = (2k+7)+2 < (k+3)^2+2\qquad$ [ Using (1)]
$= 2(k+1)+7 < k^2+6k+9+2$
$= 2(k+1)+7 < k^2+6k+11$
Now, $k^2+6k+11 < k^2+8k+16$
$\therefore 2(k+1)+7 <(k+4)^2$
$2(k+1)+7 < \{(k+1)+3 \}^2$
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., $n$.