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TurtleMatt

  • 3 years ago

Prove that each statement is true for all positive integers. 2 + 4 + 6 + ... + 2n = n^2 + n

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  1. TurtleMatt
    • 3 years ago
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    What's the first step? What do I add to both sides? I never know what to add for these questions.

  2. m_charron2
    • 3 years ago
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    I would start by simply checking what the lefthand side sums up to. Let's take smaller sums before going up to 2n. 2+4+6 = 4*3, right? 2+4+6+8=5*4. 2+4+6+8+10 = 6*5. See a trend here?

  3. TurtleMatt
    • 3 years ago
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    What do I do with that trend?

  4. freckles
    • 3 years ago
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    it is true for n=1 2=1^2+1 ... Assume it is true for n=k so we have 2+4+6+...+2k=k^2+k Now we want to show it is true for n=k+1 2+4+6+...+2k+2(k+1)=k^2+2k+2(k+1) =k^2+2k+2k+2 =k^2+2k+1+2k+1 do you think you can finish now?

  5. TurtleMatt
    • 3 years ago
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    Yeah, I can. But how did you know to add 2(k+1) to both sides? That's what confuses me.

  6. freckles
    • 3 years ago
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    We want to show it is true for n=k+1 right? so we have \[2+2(2)+2(3)+...+2(k+1)\] I replace n with k+1

  7. freckles
    • 3 years ago
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    the integer before k+1 is k

  8. freckles
    • 3 years ago
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    \[2(1)+2(2)+2(3)+...+2(k)+2(k+1)\]

  9. freckles
    • 3 years ago
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    But we know from assuming n=k we had \[2(1)+2(2)+2(3)+...+2(k)=k^2+k\]

  10. freckles
    • 3 years ago
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    So I replace 2(1)+2(3)+2(3)+...+2(k) with k^2+k in 2(1)+2(2)+2(3)+...+2(k)+2(k+1) k^2+k +2(k+1)

  11. freckles
    • 3 years ago
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    I did make a type-0 earlier but anyways this last post is good so we have k^2+k+2k+2 k^2+2k+1+k+1 (k+1)^2+(k+1)

  12. TurtleMatt
    • 3 years ago
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    Thank you!

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