Calculate sum of squares of first $n$ odd numbers Is there an analytical expression for the summation $$1^2+3^2+5^2+\cdots+(2n-1)^2,$$ and how do you derive it?

One can give a combinatorial version of the argument below. But it takes some time to tell the right story. So for now we settle for the magic math  approach. We have
$$\binom{2k+1}{3}-\binom{2k-1}{3}=\frac{(2k+1)(2k)(2k-1)-(2k-1)(2k-2)(2k-3)}{3!}.\tag{1}$$
Note that the above equation even holds when $k=1$, if we use the convention that $\binom{1}{3}=0$.
The right-hand side of (1) magically simplifies to $(2k-1)^2$.   It follows that our sum $1^2+3^2+5^2+\cdots +(2n-1)^2$ is equal to
$$ \binom{3}{3}-\binom{1}{3}   + 
 \binom{5}{3}-\binom{3}{3}  + \binom{7}{3}-\binom{5}{3}  +  \cdots + \binom{2n+1}{3}-\binom{2n-1}{3}  .$$ 
Note the wholesale cancellation. We get
$$1^2+3^2+5^2+\cdots +(2n-1)^2=\binom{2n+1}{3}.$$
