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• Module 0 Day 14 Challenge Part 1

  Fractions, Difference of Fractions

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  G

  I'm confused. The answer was n(n+1), but the options were:
  i.) 1/(n(n-2))
  ii.)1/(2n-2)
  iii.)1/(2n)
  iv.)1/(n(n-1))

  I guessed iv.) even though none of them seemed to work and the answer said: "All the denominators have the form n(n+1)." I think there was a typo in which there was a minus sign where there should have been a plus sign. Can you fix that?

• Module 0 Day 14 Challenge Part 2

  Fractions, Difference of Fractions

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  A

  @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(n-1) is the LCD; we can still algebraically set n(n-1) 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(n-1) is the LCD because the numerator becomes 1, though 🙂

  Great job!

• Module 0 Day 14 Challenge Part 3

  Fractions, Difference of Fractions

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• Module 0 Day 14 Your Turn Part 1

  Variation on a Telescoping Sum

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