Can you figure out how this math trick works?

Let i originally equal 1.

n
∑ ab+a+b = 719 //719 for this problem, anyway.
i = 1

Where a ≠ b, 1 ≤ a ≤ 5, 1 ≤ b ≤ 5, and b comes one term after a in the sequence. //Technically, order doesn’t matter, but it makes the ∑ make sense.
ab+a+b = z
Replace a and b with the new number z in the sequence and continue the summation.
i = z

So, after the first iteration, the equation becomes:

n
∑ ab+a+b = 719
i = z

1 2 3 4 5
(1*2+1+2) = 5
5 3 4 5

5*3+5+3 = 23

23 4 5

23*4+23+4 = 119

119 5
119*5+119+5 = 719

I would have to pull out my Discrete Mathematics notebook to come up with a simple formula. I’ll come back to this question when I have time. If you’re into Computer Science, think of it as a “for loop.”

As far as why order doesn’t matter, that’s kind of a trick question. The order of operations matters, so that’s why the order of the numbers from a group that you act upon doesn’t; sort of like Commutative Laws, eventually, you would get the same result no matter what two numbers you multiply and add up first in a list.

Let a#b = ab + a + b..
Then (a#b)#c = (a#b)c + a#b + c
...................... = (ab + a + b)c + (ab + a + b) + c
...................... = (ab)c + ac + bc + ab + a + b + c
...................... = a(bc) + ab + ac + a + bc + b + c
...................... = a(bc + b + c) + a + (bc + b + c)
...................... = a(b#c) + a + b#c
...................... = a#(b#c).
Since a#b = b#a, it matters neither which pair is chosen first nor in which order the operation is applied to the chosen pair.

@ratboy, I like your thinking on this.

There is, however, an additional insight that leads to a general equation. The key observation is that ab+a+b = (a+1)(b+1)-1

Using the # notation, see how this works out.
(a#b)#c =[ (a+1)(b+1)-1+1](c+1) – 1 = (a+1)(b+1)(c+1) – 1. In general, for n terms, no matter what order they are combined, you get (a1+1)(a2+1)…(an+1) – 1
For the numbers from 1 to 5, this works out to 2*3*4*5*6 -1 = 719

The mathematical concept of isomorphism provides an elegant way of stating this.

Let T(x)=x+1 Then T(a#b) = T(a)*T(b). The # operation is isomorphic to multiplication. This means that the two operators are structurally equivalent, so commutativity and associativity for # follow from the corresponding properties of multiplication. This also leads to the general formula of
T(a1#a2#...#an) = T(a1)*T(a2)*…*T(an)

@LostInParadise—nice.

Let N = {0, 1, 2, ... }, then (N, #) is a commutative monoid since 0#n = n#0 = n for each n.