Sum of all the numbers in the grid.

Question

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?

Answer 1

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 $10^2+9^2+...+1^2=385$

Answer 2

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 $i$.

Answer 3

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 $$ 2 \sum_{k = 1}^{n - 1}\sum_{j = 1}^{k} j + \sum_{j = 1}^{n} j = 2 \sum_{k = 1}^{n - 1} \frac{k(k + 1)}{2} + \sum_{j = 1}^{n} j = \frac{(n - 1) n (n + 1)}{3} + \frac{n(n + 1)}{2} = \frac{n (n + 1) (2 n + 1)}{6}. $$

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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.