Another Rectangular Grid Problem

Relevant wiki: Hockey Stick Identity

In general, the first rectangular grid consists of $1\times 2$ dots, the second of $2\times 3$ , so the ${ n }^{ th }$ consists of $n\times (n+1)$ points. Thus, the sum we are asked for is $1\times 2+2\times 3+...+n\times (n+1)$ . Note the similarity between each term of the sum and the triangular numbers formula: the ${ k }^{ th }$ term in the previous sum is $2\left( \begin{matrix} k+1 \\ 2 \end{matrix} \right)$ , therefore, the sum can be written as

$2\sum _{ k=1 }^{ n }{ \left( \begin{matrix} k+1 \\ 2 \end{matrix} \right) }$ , in which $n$ is the number of columns of the ${ n }^{ th }$ rectangular grid. This sum is equal to $2$ times the ${ n }^{ th }$ tetrahedral number: $2\left( \begin{matrix} n+2 \\ 3 \end{matrix} \right) =\frac { n(n+1)(n+2) }{ 3 }$

In the particular case $n=11$ , $\frac { 11\cdot 12\cdot 13 }{ 3 } =572$ . Hence, the junction consists of $572$ dots.

The number of dots in the $x^{th}$ grid is $x(x+1)=x^2+x$ so we are after:

$\sum_{x=1}^{11} x^2+x$

$\sum_{i=1}^{n} x=\dfrac{n(n+1)}{2},\sum_{i=1}^{n} x^2=\dfrac{n(n+1)(2n+1)}{6}$

$\sum_{x=1}^{11} x^2+x=\dfrac{11(11+1)}{2}+\dfrac{11(11+1)(2 \times 11 +1)}{6}=66+506$

$66+506=\boxed{572}$

The matrix of each grid (of n grids suppose) is of the order $n × (n + 1)$ .
Again, for more proof, the first grid has $2 × 1 = 2$ dots
The second grid has $3 × 2 = 6$ dots
The third grid has $4 × 3 = 12$ dots
The fourth grid has $5 × 4 = 20$
So, the total number of dots upto $n$ grids is $\displaystyle \sum_{n=1}^{11} n(n + 1)\\ = \boxed{572}$

I did it in an easy way. I noticed the numbers were increasing by multiples of 2, 2+4=6+6=12+8=20...etc. So I added them all (the 11 terms ) after identifying them.