<s>[INST] Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$ 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
 [/INST] 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 $
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