Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$

Question

Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$

Case $S(1)$ is true:

$$2((1)+2)\leq((1)+2)^{2}$$

$$6\leq9$$

Case $S(n)$ is true for all $n=1,2,...$

$$2(n+2)\leq(n+2)^{2}(i)$$

Case $S\left(n+1\right)$

$$2(n+3)\leq(n+3)^{2}(ii)$$

From (i)

$$2(n+2+1-1)\leq(n+2+1-1)^{2}$$

$$2(n+3)-2(1)\leq(n+3)^{2}-2(n+3)+1$$ $$2(n+3)\leq(n+3)^{2}-2(n+3)+3$$ $$2(n+3)\leq(n+3)^{2}-(2n+3)$$ $$2(2n+3)+3\leq(n+3)^{2}$$

Thus $(i)$ is true for all $n=1,2,...$

My question: How can I get the same expression as (ii)? I got to $2(2n+3)+3\leq(n+3)^{2}$ but it's clearly wrong

Answer 1

Recurrence :

$2(n+1)\leq(n+2)^2 $

For n=1

$4\leq 9$ that is correct

Suppose : $ 2(n+1)\leq(n+2)^2 $

Let's show $ 2(n+2)\leq(n+3)^2 $

$2(n+2)\leq(n+3)^2 $

$\Leftrightarrow $ $2(n+2)\leq ((n+2)+1)^2 $

$\Leftrightarrow $$2(n+2)\leq (n+2)^2 +2(n+2)+1$

$\Leftrightarrow $$0\leq(n+2)^2 +1$ that is correct

So:

$ 2(n+2)\leq(n+3)^2 $

Finally :

$ 2(n+1)\leq(n+2)^2 $

Answer 2

Note that$$2(n+2)-2(n+1)=2$$and that$$(n+3)^2-(n+2)^2=2n+5.$$So, if $2(n+1)\leqslant(n+2)^2$, you have\begin{align}2(n+2)&=2(n+1)+2\\&\leqslant(n+2)^2+2\\&\leqslant(n+2)^2+2n+5\\&=(n+3)^2.\end{align}