Pair Products Poster
Problem
Student Solutions
For 4 numbers, the first times the last is always 2 less than the product of the middle numbers.
For 5 numbers, the first times the last is always 3 less than the second times the fourth.
For $n$ numbers, the first times the last is always $n-2$ less than the second times the second-last.
Why?
Let the first number be $a$
4 numbers
$a, (a+1), (a+2), (a+3)$
First times last: $a(a+3) = a^3 + 3a$
Second times third: $(a+1)(a+2) = a^2 + 3a + 2$
Difference: $2$
5 numbers
$a, (a+1), (a+2), (a+3), (a+4)$
First times last: $a(a+4) = a^3 + 4a$
Second times fourth: $(a+1)(a+3) = a^2 + 4a + 3$
Difference: $3$
$n$ numbers
$a, (a+1), (a+2), ..., \left(a+(n-2)\right), \left(a+(n-1)\right)$
First times last: $a(a+n-1) = a^2 +an-a$
Second times second last: $(a+1)(a+n-2) = a^2+an-a+n-2$
Difference: $n-2$