Incomplete math induction Making a proof
for $n>7$
$$2^n>n^2+4n+5$$
Step 1
For $$n=7$$
$$2^7=128>7^2+4(7)+5=82$$
It is TRUE for $n=7$
Step 2
Inductive Hypotesis:  It must be true for $n=k$
$$2^k>k^2+4k+5$$
Step 3
3.1:
$$2^{k+1}>(k+1)^2+4(k+1)+5$$
But, after this it was said that...
It is needed to multiply by $2$ the expression in the induction step 2
3.2:
$$2^{k+1}>2k^2+8k+10$$
Make some transformation (that also do not understand):
3.3:
$$2^{k+1}>(k+1)^2+4(k+1)+5+k^2+2k$$
Making with that the proof by understanding that $k^2+2k>0$ for any $k\ge7$  We can deduce that $2^{k+1}>(k+1)^2+4(k+1)+5$
Questions
a: It is right to go from 2 to 3.1?  In which way?
b: How we go from 3.1 to 3.2?
c: How we go from 3.2 to 3.3?
Source: http://esaez.mat.utfsm.cl/iii.pdf, August, 2013, pp.2
 A: We want to show $2^n > n^2 + 4n + 5$ for all $7 \leq n \in \mathbb{N}$. First of all, we want to show that, this is true for $7$. Then, we will assume this is true for $n$ and try to show for $n+1$. So, step $2$ is assumption and $3.1$ is what we want to show. You wrote $3.2$ wrongly (with edits, it's now correct), but yes we will use our assumption and look what will happen when we multiply it by $2$. We get: $$2^{k+1} > 2k^2 + 8k + 10$$ Then, observe that $$(k+1)^2 + 4(k+1) + 5 + k^2 + 2k = 2k^2 + 8k + 10$$ Remember, we want to show that $2^{k+1} > (k+1)^2 + 4(k+1) + 5$. Also, $k^2+2k > 0$ . Try to finish this and if you can't please comment.
A: a. Your goal is to go from 2 to 3.1; that will complete the induction.  
b,c. You do not go from 3.1 to anywhere, you end up at 3.1.
Here is how to go from 2 to 3.1:
Method 1: Multiply both sides of step 2 by $2^1$.  This will make the LHS agree with 3.1.  Then you will need to prove that $2(k^2+4k+5) > (k+1)^2+4(k+1)+5$, to make the RHS agree.
Method 2: Add a polynomial in $k$ to both sides of step 2, to make the RHS agree with 3.1.  The correct polynomial is $(k+1)^2+4(k+1)+5-(k^2+4k+5)=2k+5$.  Then you will need to prove that $2^{k+1}>2^k+(2k+5)$, to make the LHS agree.
