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Pssssst

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)

  • one year ago
  • one year ago

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  1. Pssssst
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    @ParthKohli

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

    • one year ago
  3. ParthKohli
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    Can you find the relationship between those two?

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

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

    • one year ago
  6. Spacelimbus
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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=

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

    • one year ago
  8. ParthKohli
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    But 8 + 16 = 24

    • one year ago
  9. ParthKohli
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    It is a series, not a sequence.

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

    • one year ago
  11. ParthKohli
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    No problem!

    • one year ago
  12. Spacelimbus
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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.

    • one year ago
  13. ParthKohli
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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.

    • one year ago
  14. tcarroll010
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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.

    • one year ago
  15. ParthKohli
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    ^ that is simple induction, correct.

    • one year ago
  16. tcarroll010
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    All good now @Pssssst ?

    • one year ago
  17. tcarroll010
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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.

    • one year ago
  18. Pssssst
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    Thank you, everyone! Helped a lot :)

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

    • one year ago
  20. Pssssst
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    @tcarroll010 Do you know much about history?

    • one year ago
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