Reformulating a sum I´m struggling to disentangle this sum... Stuck in getting from line 2 to 3. Could someone give me a hint what the rationale behind this is? In particular, why are we squaring inside the summation here?
$$\sum_{t=2}^\tau G_t=\sum_{t=2}^\tau \Delta_t\sum_{i=1}^{t-1}\Delta_i$$
$$=\sum_{2\le t\le \tau,i< t} \Delta_t\Delta_i$$
$$2\sum_{t=2}^\tau G_t =\left(\sum_{t=1}^\tau \Delta_t\right)^2-\sum_{t=1}^\tau {\Delta_t}^2$$
I did not understand the last change of summation.
Thank you!
 A: Write the equation as
$$
\eqalign{
  & \sum\limits_{2\, \le \,t\, \le \,\tau } {G(t)}  = \sum\limits_{2\, \le \,t\, \le \,\tau } {\Delta (t)\sum\limits_{1\, \le \,i\, \le \,t - 1} {\Delta (i)} }  =   \cr 
  &  = \sum\limits_{1\, \le \,i\, < \,t\, \le \,\tau } {\Delta (t)\Delta (i)}  \cr} 
$$
understanding that the summation is over the inidices $i$ and $t$
Inverting the indices, you get the same sum
$$
\sum\limits_{2\, \le \,t\, \le \,\tau } {G(t)}  = \sum\limits_{1\, \le \,t\, < \,i\, \le \,\tau } {\Delta (t)\Delta (i)} 
$$
Add the two together
$$
\eqalign{
  & 2\sum\limits_{2\, \le \,t\, \le \,\tau } {G(t)}
  = \sum\limits_{1\, \le \,i\, < \,t\, \le \,\tau } {\Delta (t)\Delta (i)}  + \sum\limits_{1\, \le \,t\, < \,i\, \le \,\tau } {\Delta (t)\Delta (i)}  =   \cr 
  &  = \sum\limits_{1\, \le \,i\, \ne \,t\, \le \,\tau } {\Delta (t)\Delta (i)}  = \sum\limits_{\matrix{
   {1\, \le \,i\, \le \,\tau }  \cr 
   {1\, \le \,t\, \le \,\tau }  \cr 
 } } {\Delta (t)\Delta (i)}  - \sum\limits_{1\, \le \,i\, = \,t\, \le \,\tau } {\Delta (t)\Delta (i)}  =   \cr 
  &  = \left( {\sum\limits_{1\, \le \,t\, \le \,\tau } {\Delta (t)} \sum\limits_{1\, \le \,i\, \le \,\tau } {\Delta (i)} } \right)
 - \sum\limits_{1\, \le \,\,t\, \le \,\tau } {\Delta (t)\Delta (t)}  =   \cr 
  &  = \left( {\sum\limits_{1\, \le \,t\, \le \,\tau } {\Delta (t)} } \right)^{\,2}  - \sum\limits_{1\, \le \,\,t\, \le \,\tau } {\Delta (t)^{\,2} }  \cr} 
$$
