Groups of numbers arranged as odd numbers.

Question

I came across arrangement of numbers grouping them with a table of three, five, seven, eleven etc... for example:

:$1----2----3$

:$4----5----6$

:$7----8----9$

:$10---11---12$

:$13---14---15$

:$16---17---18$

:$19---20---21$

$2\times3=6$, $5\times3=15$, $8\times3=24$ so for each middle number in a given row, multiplication by $3$ gives the sum of the three numbers in that row. Summing all middle numbers and multiplying by $3$ gives the sum of all $21$ numbers.

The same properties apply to numbers arranged in a group of seven.

:$1----2----3----4----5----6----7$

:$8----9---10----11---12---13---14$

:$1---16---17----18---19---20---21$

$4$ being the middle in the first row gives $4\times7=28$, sum of the numbers in the first row. Similarly $11\times7=77$ and $18\times7=126$ and also $28+77+126=231$, the sum of all $21$ numbers. $11$ being the center of the groups, times $7$ times $3$ equals $231$.

Could you give me any explanation as to why this method of calculation works?

Answer 1

Think of the numbers above as a matrix. The first above is $7\times3$ and the second one above is $3\times 7$.

Start thinking about summing all numbers is a given row. Either you sum all the numbers. Or you multiply the middle number by the number of columns in the matrix. The reason why this gives you the same answer is that, for row with $3$, say, columns, $\begin{array}{ccc}n-1&n&n+1\end{array}$, the sum is $3n$. For $5$ columns you have $\begin{array}{ccccc}n-2&n-1&n&n+1&n+2\end{array}$, which sums to $5n$.

Notice that this observation allows you to construct similar procedure for arrangements, matrices, with even number of columns. For, $4$, say, $\begin{array}{cccc}n-2&n-1&n&n+1\end{array}$, which sums to $4n-2$, or, alternatively, $\begin{array}{cccc}n-1&n&n+1&n+2\end{array}$, which sums to $4n+2$.

Yet another rule you can create is not to multiply the middle number, but, say, the first number. That is, instead of $\begin{array}{ccc}n-1&n&n+1\end{array}$, label things as $\begin{array}{ccc}n&n+1&n+2\end{array}$. The rest is limited only by your immagination.