Re: Add the numbers in a 9x9 multiplication Table

On Mon, 9 Jun 2025 16:32:37 -0000 (UTC), Richard Tobin wrote:

   sum(x=1..n) sum(y=1..n) [xy]
=  sum(x=1..n) [x sum(y=1..n) y]
=  [sum(x=1..n) x] [sum(y=1..n) y]
=  [sum(x=1..n)]^2
 
sum(x=1..n) is well known and easily seen geometrically.

Well blow me down with a feather.

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To calculate the sum of the numbers 1 through n:

1, 2, 3 ... n-2, n-1, n

We can write the series in reverse:

n, n-1, n-2 ... 3, 2, 1

Adding the two rows together each element provides the same sum, (n+1).
so:

2[sum(x=1..n)] = n(n+1)

and

[sum(x=1..n)] = n(n+1)/2

   sum(x=1..n) sum(y=1..n) [xy]
=  sum(x=1..n) [x sum(y=1..n) y]
=  [sum(x=1..n) x] [sum(y=1..n) y]
=  [sum(x=1..n)]^2

The value we are looking for, the sum of the elements in a n x n
multiplication table, is given by

[sum(x=1..n)]^2 = [n(n+1)/2]^2

= (n^4 + 2n^3 + n^2)/4

Thanks, that's brill.
--
David Entwistle