Proof by induction for $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$ for $k > 4$

Question

I was given this proof for hw.

Prove that $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$

So, far I've gotten this

Basis:
$k = 5$, $2^{5 + 1} - 1 > 2\cdot5^2 + 2\cdot5 + 1$ => $63 > 61$ (So, the basis holds true)

Hypothesis:
for all $k > 4$, $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$

Inductive step:
LHS $2^{k + 1 + 1} - 1 = 2^{k + 2} = 2 \cdot 2^{k + 1}$

$2[ 2^{k + 1} - 1 + 1] - 1 \geqslant 2\cdot(2k^2 + 2k + 1 + 1) - 1$

RHS (Attempt to prove that the RHS is less than the LHS )

$2(2k^2 + 2k + 1 + 1) - 1 \geqslant 2(k + 1)^2 + 2(k + 1) + 1$

Next step
$2(k + 1)^2 + 2(k + 1) + 1 = 2k^2 + 4k + 2 + 2k + 1 + 1 = (2k^2 + 2k + 1) + 4k + 3$

Now I'm stuck at how to prove that
$2(2k^2 + 2k + 1 + 1) - 1 \geqslant (2k^2 + 2k + 1) + 4k + 3$

So, what can I do?

Thanks

Answer 1

Hypothesis: $$\begin{align} \forall k > 4,&\quad\; 2^{k + 1} - 1 > 2(k)^2 + 2(k) + 1 \\ &\iff 2^{k+1} \gt 2k^2 + 2k + 2 \end{align}$$

Inductive step: $$\begin{align} 2^{k + 1 + 1}& = 2\cdot 2^{k+1}\\ & \gt 2\cdot (2k^2 + 2k+2) \\ & = 4k^2 + 4k + 4\\ & = (2k^2 + 4k + 2) + (2k^2 + 2)\\ & = 2(k+1)^2 + 2k^2 + 2 \\ & \geq 2(k+1)^2 + 2(k + 1) + 2\tag{for $k \geq 2$} \end{align}$$

Answer 2

Inductive step

$$2^{k+2}-1=2\times 2^{k+1}-1>2(2k^2+2k+2)-1>2(k+1)^2+2(k+1)+1$$