Question

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

Answer 1

Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):(2n+7)<{(n+3)}^2$
$P\left(n\right)$ is true for $n=1$ since $2.1+7=9<{(1+3)}^2=16$, which is true.
Let $P\left(k\right)$ be true for some positive integer $k$, i.e.,
$(2k+7)<{(k+3)}^2........\left(i\right)$
We shall now prove that $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Consider
$2(k+1)+7=\left\{\right(2k+7)+2\}<{(k+3)}^2+2$ [Using $\left(i\right)$]
$2(k+1)+7<k^2+6k+9+2$
$2(k+1)+7<k^2+6k+11$
Now, $k^2+6k+11<k^2+8k+16$
$\therefore$ $2(k+1)+7<{(k+4)}^2$
$2(k+1)+7<{\left\{\right(k+1)+3\}}^2$
Thus, $P(k+1)$ is true whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all natural numbers i.e., $n$.