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    How does step 1 convert to step 2? We don't know what n is so I think step 2 is just making assumptions

    Module 1 Day 1 Your Turn Part 1
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    • The Blade DancerT
      The Blade Dancer M0★ M1★ M2★ M3★ M4 M5
      last edited by debbie

      Module 1 Week 1 Day 1 Your Turn Part 1 Mini-Question

      478409c8-b8b7-4490-80be-81d16233363c-image.png

      How does step 1 convert to step 2? We don't know what n is so I think step 2 is just making assumptions

      The Blade Dancer
      League of Legends, Valorant: Harlem Charades (#NA1)
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      debbieD 1 Reply Last reply Reply Quote 1
      • debbieD
        debbie ADMIN M0★ M1 M5 @The Blade Dancer
        last edited by debbie

        @the-blade-dancer The second step looks quite different from the first step! It's because for the first two fractions, we have multiplied the top and bottom by a factor.

        For the first fraction, we have multiplied the top and bottom by \( n + 1.\)

        \(\frac{1}{\textcolor{red}{n}} = \frac{n + 1}{\textcolor{red}{n} (n + 1) } \)

        For the second fraction, we have multiplied the top and bottom by \(n.\)

        \(\frac{1}{\textcolor{red}{n+1}} = \frac{n}{\textcolor{red}{(n +1)} n } \)

        For the third fraction, we've done something different: we have factored the bottom. (Factor just means to write as a product of two terms, like \( 12 = 3 \times 4.\) )

        \( - \frac{1}{n^2 + n} = - \frac{1}{n(n+1)} \)

        Why are we trying to do this? To make the bottoms of the fractions all equal to \( n(n+1).\) Then we can add the fractions together very easily, and luckily we have some terms cancel out, so we get \(0\) on the top, which is a little fortunate!

        \( \frac{n + 1 - n - 1}{n(n+1)} \)

        \( = \frac{0}{n(n+1)} = \boxed{0}.\)

        🙂

        high-five.png

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