$\sum_{r=0}^m(2r-1)\sum_{r=0}^m\,^mC_r\sum_{r=0}^m1$ to $m^2-1\;2^m\;m+1$ 
$$\begin{align}
\sum_{r=0}^m\Delta_r&=
\begin{vmatrix}
\displaystyle\sum_{r=0}^m(2r-1)&\displaystyle\sum_{r=0}^m\,^mC_r&\displaystyle\sum_{r=0}^m1\\
m^2-1&2^m&m+1\\
\sin^2\left(m^2\right)&\sin^2(m)&\sin^2(m+1)
\end{vmatrix}\\
&=\begin{vmatrix}
m^2-1&2^m&m+1\\
m^2-1&2^m&m+1\\
\sin^2\left(m^2\right)&\sin^2(m)&\sin^2(m+1)
\end{vmatrix}=0
\end{align}$$

I don't understand how they have gone from the first row in the top determinant to the first row in the second determinant.
 A: Hints:


*

*$\sum_{r=0}^m(2r-1)=\frac{(-1+2m-1)(m+1)}{2}=m^2-1$

*Do you know that $C_n^0 + C_n^1+\cdots +C_n^n=2^n$
A: Actually, I do not think your difficulty is with the determinant, but rather with how the various summations shake themselves out.    
$\sum_{r=0}^{m}(2r-1)=    
\sum_{r=0}^{0}(2r-1)+\sum_{r=1}^{m}(2r-1)=    
-1+\sum_{r=1}^{m}(2r-1)$    
You do know, I hope, that the sum of the first $m$ odd positive numbers is $m^2$. In any event, this is provable via mathematical induction. So much for column 1.     
The sum of the binomial coëfficients for a given power is always a power of $2$. The apex of the Pascal triangle is the number 1. To progress from one row of the triangle to the next, each member of the current row is represented twice in the succeeding row, so the sum of the members in each row doubles from one row to the next. Trivially, the number 1 is a power of 2, so all succeeding rows resulting from such doubling will be, also.   
$\begin{matrix}
[r=0]&1&=1=2^0\\
[r=1]&1+1&=2=2^1\\
[r=2]&1+2+1&=4=2^2\\
[r=3]&1+3+3+1&=8=2^3\\
[r=4]&1+4+6+4+1&=16=2^4\\
\end{matrix}$, etc.
So much for column 2.    
The third column is the sum of $m$ units (1s) plus another unit since the starting point was $0$.
