Is this a correct way to prove by induction? I understand that most people use the inductive hypothesis, but I find that counterintuitive.  Is the below proof correct?  In particular, I am concerned with my use of $n$ in (I); is the reason people use another variable, e.g. $k$, for conceptual reasons, or does the use of $n$ create an error in my proof? 
\begin{align*}
 P(0) \land [P(k) \Rightarrow P(k+1)] \implies P(n) \tag{AI} \\
\end{align*}

\begin{align*}
\text{When }n = 2, \ \ 2 + 6 + 10 + . . . + (4n - 2) &= 2n^2 \tag{B} \\
2 + 6 &= 2 \cdot 2^2 \\
8 &= 8 \\
\end{align*}

$$
\begin{pmatrix}
 2 + 6 + 10 + ... + (4n-2) = 2n^2 \\
 \Big\Downarrow \\
 2 + 6 + 10 + ... + (4n-2) + (4(n+1)-2) = 2(n+1)^2 \\
\end{pmatrix} \tag{I}
$$
\begin{align*}
\Big\Updownarrow
\end{align*}
\begin{align*}
4(n+1)-2 &= 2(n+1)^2 -2n^2 \\
4n+4-2 &= 2n^2+4n+2 - 2n^2 \\
4n+2 &= 4n+2 \\
\end{align*}

$$
 \text{B} \land \text{I} \land \text{AI} \implies 2 + 6 + 10 + . . . + (4n - 2) = 2n^2 \text{ for } n > 1 \ \ \square
$$

Is this alternate solution correct?  I am confident in my reasoning, but am unsure if it is a valid mathematical argument.
Pairing $1$st with $n$th term, $2$nd with $(n-1)$th term, etc., yields $\mathbf{\frac{n}{2}}$ pairs:
\begin{align*}
 2 + (4n-2) \ \ + \ \ 6 + (4(n-1)-2) \ \ + \ \ 10 + (4(n-2)-2) \ \ + \ \ ... &= 2n^2 \\
 2 + (4n-2) \ \ + \ \ 6 + (4n-6) \ \ + \ \ 10 + (4n-10) \ \ + \ \ ... &= 2n^2 \\
 4n \ \ + \ \ 4n \ \ + \ \ 4n \ \ + \ \ ... &= 2n^2 \\
 4n \cdot \mathbf{\frac{n}{2}} &= 2n^2 \\
 2n^2 &= 2n^2 \\
 &\ \square \\\ \\
\end{align*}

The first proof is perfectly fine. The choice of letters is mostly/often just for aestetical/pedagogical reasons.
The second (and to some extent the first) uses the $\dots$ informal notation. This should probably be seen as a shorthand or more visual way of handling summations. Formally one should have used $\sum$-notation. The problem in the last is that you visually rearranges the terms which is not that good looking when using $\sum$ notation. 
The second approach if done strictly would also need induction to be complete (unless you rely on propositions that already have been proven by induction).
