<s>[INST] How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that  
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$. 
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI. 
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the
formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$
$\diamond$
 [/INST] Induction is fundamental to the structure of mathematics. The $``\cdots"$ in the expression $1+ 5+ 9 + \cdots +(4n-3)$ cannot be explained rigorously without using some form of induction. All of the picture proofs here have an implied $``\cdots"$ in them. Similarly, $\text{$\Sigma$-notation}$ and telescoping sums cannot be defined without some form of induction. 
Putting that aside, I have always been curious about the process of converting sums to integrals, so I offer the following flawed solution.
Note that
$\int_\limits k^{k+1}(4x-5)\,dx 
 = \left. (2x^2-5x) \right|_k^{k+1}
 = (2k^2+4k+2-5k-5)-(2k^2-5k)
 = 4k-3$.
So
\begin{align}
    \sum\limits_{k = 1}^n (4k - 3)
    &= \sum\limits_{k = 1}^n \left( \int_\limits k^{k+1}(4x-5)\,dx \right) \\
    &= \int_\limits 1^{n+1}(4x-5)\,dx 
        &\text{(This step requires induction.)} \\
    &= \left. (2x^2-5x) \right|_1^{n+1} \\
    &= (2n^2 + 4n+2 -5n - 5) - (2 - 5) \\
    &= 2n^2-n
\end{align}
Addendum
I got the integrand by solving 
$4k - 3 = \int_\limits k^{k+1}(2ax+b)\,dx = 2ak + (a+b)$
The implication is that
\begin{align}
    1 &= u_2 - u_1 \\
    5 &= u_3 - u_2 \\
    9 &= u_4 - u_3 \\
    &\vdots \\
    4k-3 &= u_{k+1} - u_k \\
    &\vdots
\end{align}
where $u_k = 2k^2 - 5k$.
Which converts the sum to the sum of a collapsing series.
This implies Lynn's graphical proof shown below and, though it still requires induction, I thought it shows an interesting way to compute sums.
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