Sum of all the numbers in the grid. 
A square containing numbers
  $$
\begin{array}{|c|c|c|}
\hline 1 & 2 & 3 \\ \hline
1 & 2 & 2 \\ \hline
1 & 1 & 1 \\\hline
\end{array}
\qquad \qquad\qquad
\begin{array}{|c|c|c|c|}
\hline 1 & 2 & 3 & 4 \\ \hline
1 & 2 & 3 & 3 \\ \hline
1 & 2 & 2 & 2 \\ \hline
1 & 1 & 1 & 1 \\\hline
\end{array}
$$
  Continue this pattern until the box is $10 \times 10$.
  Then add all the numbers together.

So this was my attempt.
Set the square $1\times1$ as $a_1$ square $2 \times 2$ as $a_2$ and so on. $a_1$ to $a_2$'s d is $(1 \cdot 2)+2 a_2$ to $a_3$'s d is $(1 \cdot 2)+(2 \cdot 2)+3$ and so on. So then I can the individual ds from each square to the next square. 
And because
$$
1+(1+2)+(1+2+3)\ldots+(1+2+3+\ldots+10)
= \sum_{k=1}^{1} 1 + \sum_{k=1}^{2} 1+ \ldots +\sum_{k=1}^{10} 1=\sum_{k=1}^{10} \frac{n(n+1)}{2}
$$
so
$$
(1\cdot2)+(1\cdot2)+(2\cdot2)\ldots+(1\cdot2)+(2\cdot2)\ldots+(10\cdot2)
= \sum_{k=1}^{1} 2k + \sum_{k=1}^{2} 2k+\ldots+\sum_{k=1}^{10} 2k
$$
which also means
$$
\sum_{k=1}^{10} 2 \cdot \frac{n(n+1)}{2}
=\sum_{k=1}^{10} {n^2+n}
$$
And these are my calculations 
$$\frac{10 \cdot 11 \cdot 21}{6}+\frac{10 \cdot 11}{2}$$
$$385+55=440$$
$$1+2+3+4+\ldots+10=55$$
$$440+55=495$$
It all seems right to me but the answer says its 385 is there any thing I did wrong?

To be honest I am not sure what you are doing. Let me add my solution. One way to write down a formula for the $n \times n$ version is $$\sum_{i = 1}^n i(2(n+1-i) - 1).$$
You will find $2n-1$ times the number $1$, $2(n-1)-1$ times the number $2$ etc. This means we want to add these up and then also add the values to the numbers by multiplying with $i$.
