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

Let $P(n)$ be the statement $2n+7<(n+3)^2$.

Base case: For $n=1$,

Since $9<16$, $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$:

We must show $2(k+1)+7<(k+4)^2$, i.e. $2k+9<(k+4)^2$.

Adding $2$ to both sides of the hypothesis:

Now compare $(k+3)^2+2$ with $(k+4)^2$ using the difference of squares:

Since $k\ge1$ (a natural number), $2k+7\ge9>2$, so …

Inequality induction. Add $2$ to the induction hypothesis, then use the difference of squares $(k+4)^2-(k+3)^2=2k+7$ together with $2k+7>2$ (true since $k\ge1 …

• Adding $2$ to the hypothesis but then not proving a further inequality is needed — assuming $(k+3)^2+2=(k+4)^2$ without checking (in fact the difference is $2k+7$, which merely needs to exceed $2$, not equal it).
• Sign/arithmetic slip in the difference of squares expansion $(k+4)^2-(k+3)^2$. …