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?