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Pssssst

  • 2 years ago

SOMEONE PLEASE HELP I'M ALMOST DONE. Use mathematical induction to prove the statement is true for all positive integers n. 8 + 16 + 24 + ... + 8n = 4n(n + 1)

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  1. Pssssst
    • 2 years ago
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    @ParthKohli

  2. ParthKohli
    • 2 years ago
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    True for the base case. Now assume that it is true for k, and then prove for k + 1.

  3. ParthKohli
    • 2 years ago
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    Can you find the relationship between those two?

  4. Spacelimbus
    • 2 years ago
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    Is it supposed to be \[\Large 4n(n+1)  \] If that's the case, it must run a bit strangely. \[n=1 \longrightarrow 8 \checkmark \\n=2 \longrightarrow24 \neg\checkmark \]

  5. ParthKohli
    • 2 years ago
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    Actually, the order of operations first tell you to multiply 4 and \(n\), then \(n +1\)

  6. Spacelimbus
    • 2 years ago
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    I believe I can't follow you @ParthKohli \[\Large 4n(n+1)=4n^2+4n \] right? and still n=1 -> 8 n=2 -> 24, or where do I miss something? (which is indeed possible=

  7. ParthKohli
    • 2 years ago
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    The question is actually saying \((4 \cdot n) \cdot (n + 1)\)

  8. ParthKohli
    • 2 years ago
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    But 8 + 16 = 24

  9. ParthKohli
    • 2 years ago
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    It is a series, not a sequence.

  10. Spacelimbus
    • 2 years ago
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    oh I understand, this is where I misunderstood the problem, thank you.

  11. ParthKohli
    • 2 years ago
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    No problem!

  12. Spacelimbus
    • 2 years ago
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    It's not explicit mentioned in the problem set, I have seen such problems as mainly sequences/progressions, in which this would be arithmetic.

  13. ParthKohli
    • 2 years ago
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    Spacelimbus, can you please help the asker? I wish I could, but I am currently on a mobile device and it is hard for me to type.

  14. tcarroll010
    • 2 years ago
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    Now that you know it works for the base case, you assume it holds for the case of "k", and you try to prove it works for "k + 1". So, add 8(k + 1) to both sides: 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k(k + 1) + 8(k + 1) 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k^2 + 4k + 8k + 8 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4k^2 + 12k + 8 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k^2 + 3k + 2) 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k + 1)(k + 2) 8 + 16 + 24 + ... + 8k + 8(k + 1) = 4(k + 1)[(k + 1) + 1] But that is just the same expression we assumed true, but now holding for "k + 1", so this proves the case.

  15. ParthKohli
    • 2 years ago
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    ^ that is simple induction, correct.

  16. tcarroll010
    • 2 years ago
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    All good now @Pssssst ?

  17. tcarroll010
    • 2 years ago
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    The reason this proves the case is because of the way mathematical induction works. The base case has already been shown to be true. Our first "k" is "1" and we have proved it for "k + 1" or "2". "2" becomes our new "k", and by induction, it is proved for "k + 1" or "3", and this process goes on indefinitely. That is why mathematical induction works and why the above proof works.

  18. Pssssst
    • 2 years ago
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    Thank you, everyone! Helped a lot :)

  19. tcarroll010
    • 2 years ago
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    Glad we were able to help. Thanks for the recognition!

  20. Pssssst
    • 2 years ago
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    @tcarroll010 Do you know much about history?

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