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?
 A: Consider the L-shaped figure $L_i$. $L_1$ adds $1.19$, $L_2$ adds $2.17$, $L_3$ adds $3.15$ and so on till $L_{10}$ adds $10.1$ for a total of $385$. $1.19+2.17+3.15+...+10.1=385$
Edit: Another way is to look at it as a 3d pyramid where each cube adds a $1$ to the sum. The ground floor has $10^2$ cubes, first floor has $9^2$ cubes and so on giving $10^2+9^2+...+1^2=385$ 
A: 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 $i$.
A: Another way to think of it is by summing diagonals: A $n \times n$ square has the following diagonal sums
$$
1, \\ 1 + 2 , \\ \vdots \\ 1 + \ldots + n - 1, \\ 1 + \ldots + n, \\ 1 + \ldots + n - 1, \\ \vdots \\ 1 + 2, \\ 1.
$$
Therefore, the formula for the sum is
$$
2 \sum_{k = 1}^{n - 1}\sum_{j = 1}^{k} j + \sum_{j = 1}^{n} j
= 2 \sum_{k = 1}^{n - 1} \frac{k(k + 1)}{2} + \sum_{j = 1}^{n} j 
= \frac{(n - 1) n (n + 1)}{3} + \frac{n(n + 1)}{2}
= \frac{n (n + 1) (2 n + 1)}{6}.
$$

We notice that this is also the formula for the sum of the first $n$ square numbers. This is because if you consider the sum of all entries of the and mirrored $L$-shape (not counting the one element that is in the row and column simultaneously twice) you get
$$
\sum_{j = 1}^{k} j  + \sum_{j = 1}^{k - 1} j
= \frac{k(k + 1)}{2} + \frac{k(k - 1)}{2}
= \frac{k^2 + k + k^2 - k}{2}
= k^2.
$$
To explain the $L$-shape better: In the first square the first L is just the lower left number, one. The next $L$ consist of the adjacent numbers, $1 + 2 + 1 = 2^2$ and the next one captures the remaining ones.
