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 $n &gt; 0$. 

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.

$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
&amp; = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
&amp; = (2k^2 − k) + (4k + 1) \\
&amp; = 2k^2 + 3k + 1 \\
&amp; = 2(k + 1)^2 − (k + 1)
\end{align}
$$

$\diamond$
Follow the rainbow!

$$ \Large \color{PaleVioletRed}1 + \color{DarkViolet}5 + \color{DodgerBlue}9 + \dots + \color{LightCoral}{(4n-3)} = n(2n-1) = 2n^2-n$$
One way using distributive laws and multiplication:

$\sum_{k=1}^n(4k-3)$

$=\sum_{k=1}^n(4k)+\sum_{k=1}^n(-3)$

$\sum_{k=1}^n(-3)=-3n$

$\sum_{k=1}^n(4k)=4\cdot\sum_{k=1}^n(k)$

Assuming n is even:

$\sum_{k=1}^n(k)=1+2+3+4...+n=(1+n)+(2+n-1)...+(\frac{n}{2}+\frac{n}{2}+1)=\frac{n(1+n)}{2}$

Assuming n is odd:
$\sum_{k=1}^n(k)=n+\sum_{k=1}^{n-1}k=\frac{2n}{2}+\frac{(n-1)(n)}{2}=\frac{n(1+n)}{2}$

Therefore it works for all n

Putting all of these together:$\sum_{k=1}^n(4k-3)=4\cdot\sum_{k=1}^n(k)+\sum_{k=1}^n(-3)=\frac{4n(1+n)}{2}-3n=2n^2+2n-3n=2n^2-n$
Q.E.D. 
Or you could just know the summation identities and easily proof it.