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egbeach

  • one year ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 1. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) =

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  1. geerky42
    • one year ago
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    4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) \(\textbf=\) ??? what?

  2. egbeach
    • one year ago
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    it doesnt say... thats how it was written

  3. zzr0ck3r
    • one year ago
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    ouch

  4. zzr0ck3r
    • one year ago
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    it is not a statement...

  5. geerky42
    • one year ago
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    we cannot prove expression "4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2)"

  6. geerky42
    • one year ago
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    I mean it doesn't make sense.

  7. egbeach
    • one year ago
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    thats what i thought

  8. zzr0ck3r
    • one year ago
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    I cant find the formula elsewhere.

  9. geerky42
    • one year ago
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    What I realize though is that pattern is supposed to be \(n(n+2)\) starting at \(n=4\), not \(4n(4n+2)\) Because otherwise pattern is supposed to be \(4\cdot6+8\cdot10+12\cdot14+\cdots\)

  10. egbeach
    • one year ago
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    okay i found it. 4 ⋅ 6 + 5 ⋅ 7 + 6 ⋅ 8 + ... + 4n( 4n + 2) = (4(4n+1)(8n+7))/6

  11. zzr0ck3r
    • one year ago
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    right @geerky42

  12. zzr0ck3r
    • one year ago
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    Do you want to see if it matches the formula she found?

  13. geerky42
    • one year ago
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    Maybe I misinterpreted the pattern, it's just \(\displaystyle \sum_{i=4}^{4n}i(i+2)\) ?

  14. egbeach
    • one year ago
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    wait i found the answer since the statement is false. thank you though

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