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sasogeek

  • 3 years ago

prove by induction that \(\large 1 \times 2 + 2 \times 3 + ... +n(n+1) = \frac{1}{3}n(n+1)(n+2) \)

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  1. Mimi_x3
    • 3 years ago
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    well, where are you stuck i assume you should know the steps n=k, then n=k+1

  2. Mimi_x3
    • 3 years ago
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    Assume \(n=k\) is true \[1*2+2*3+...+k(k+1) = \frac{1k}{3}(k+1)(k+2)\] Prove \(n=k+1\) \[1*2+2*3+...+k(k+1)+(k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}\] \[\frac{(k)(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}\] Now, you can prove RHS = LHS

  3. sasogeek
    • 3 years ago
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    ok i understand and have worked the up to the point \(\large 1* 2 + 1* 3 + ... + k+1(k+2) = 1 * 2 + 1*3 + ... +k(k+1) +k+1(k+2) \) \(\large = \frac{1}{3}k(k+1)(k+2)+(k+1)(k+2)\) then what?

  4. RadEn
    • 3 years ago
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    for left side : k(k+1)(k+2)/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1) = (k+1)(k+2)(k+3)/3 same like right side

  5. Mimi_x3
    • 3 years ago
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    Well, you prove as I said above. Prove the LHS = RHS \[=>(k+1)(k+2)\left[ \frac{k}{3}+1\right]\]It should be straight forward now.

  6. Mimi_x3
    • 3 years ago
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    Just, some algebra and you're done!

  7. sasogeek
    • 3 years ago
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    it's still blurry but i'll try to get it in a bit :)

  8. Mimi_x3
    • 3 years ago
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    Well, where are you stuck?

  9. Mimi_x3
    • 3 years ago
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    All you have to do here is: Prove the LHS that is: \[\frac{(k)(k+1)(k+2)}{3} + (k+1)(k+2) \] Is equal to the RHS: \[=> \frac{(k+1)(k+2)(k+3)}{3}\]

  10. sasogeek
    • 3 years ago
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    ohh, thanks :) normal algebra takes off from there i see :) makes sense now xD

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