Algebra for Life

Module 0 Day 14 Challenge Part 2
I think you can prove this pattern algebraically: here are my thoughts. (1/n1)  1/n where n and n1 are consecutive = 1/ (n1) n. The LCD(Least Common Denominator) is (n1)n since consecutive numbers are coprime. then, setting the two fractions to their common denominators you get [n(n1)] / (n1)n = 1/(n1)n therfore, DONE

@angeliccanary Yep, that's exactly how you would rigorously prove that a pattern exists!
Just a really small note for the proof, it actually doesn't strictly matter that n(n1) is the LCD; we can still algebraically set n(n1) to be the common denominator no matter what. This is sort of like how we can write 1/2 + 3/5 as 22/20, even though technically it's 11/10 simplified. You're right that n(n1) is the LCD because the numerator becomes 1, though
Great job!