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tux

  • 3 years ago

Prove by induction: http://i.imgur.com/715S5.png

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  1. mukushla
    • 3 years ago
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    Proof: For each positive integer \(n\ge2\) , let \(S(n)\) be the statement\[1.2+2.3+...+(n-1)n=\frac{n(n-1)(n+1)}{3}\] Basis step: S(2) is the statement \(1.2=\frac{2.1.3}{3}=2\). Thus \(S(2)\) is true. Inductive step: We suppose that \(S(k)\) is true and prove that \(S(k+1)\) is true. Thus, we assume that \[1.2+2.3+...+(k-1)k=\frac{k(k-1)(k+1)}{3}\] and prove that \[1.2+2.3+...+k(k+1)=\frac{k(k+1)(k+2)}{3}\] what to do now?

  2. mukushla
    • 3 years ago
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    @tux make sense?

  3. tux
    • 3 years ago
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    You rewritten sum as \[\sum_{i=1}^{n-1}i(i+1)=\sum_{i=1}^{n}i(i+1)+(n-1)(n-1+1)\] ? And then substituted n=k in induction step Then n=k+1 you got ((k+1)-1)(k+1) k*(k+1)=\[\frac{ k(k+1)(k+2) }{ 3 }\]

  4. mukushla
    • 3 years ago
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    this is how we got things done by induction We assume that \(S(k)\) is true and we want to prove that \(S(k+1)\) is true.

  5. tux
    • 3 years ago
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    Thank you. Now I can do it alone

  6. mukushla
    • 3 years ago
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    yw :)

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