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anonymous

  • one year ago

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. anonymous
    • one year ago
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    @Shalante ?

  2. ganeshie8
    • one year ago
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    Is the statement true for \(n=1\) ?

  3. anonymous
    • one year ago
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    yes

  4. ganeshie8
    • one year ago
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    how, can u explain

  5. anonymous
    • one year ago
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    4(1)(1+1) 4(2) =8

  6. ganeshie8
    • one year ago
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    that means the expression \(4n(n+1)\) correctly gives the sum on left hand side for \(n=1\)

  7. ganeshie8
    • one year ago
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    what if you have 2 terms on left hand side : \(8+16\) is the statement still true ?

  8. anonymous
    • one year ago
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    wait... you lost me. i have to plug in both now?

  9. ganeshie8
    • one year ago
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    you don't need to, one base case is sufficient here, do you know how induction works ?

  10. anonymous
    • one year ago
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    i'm a little rusty on induction.

  11. ganeshie8
    • one year ago
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    basically your goal is to prove that the given statement is true for all positive integers : \(n=1,2,3,\ldots \)

  12. anonymous
    • one year ago
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    i understand that. but I get confused after I've already proven 1

  13. anonymous
    • one year ago
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    i don't know where to go from there

  14. ganeshie8
    • one year ago
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    suppose, you "know" that the given statement is true for \(n=k+1\), whenever it is true for \(n=k\)

  15. ganeshie8
    • one year ago
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    is that sufficient to claim that the given statement is true for all integers : \(n=1,2,3,\ldots\) ?

  16. anonymous
    • one year ago
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    so then i have to plug in k+1

  17. ganeshie8
    • one year ago
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    forget about the process, il surely give you the complete proof later for now, im just asking you a question..

  18. ganeshie8
    • one year ago
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    Here it is again : suppose, you "know" that the given statement is true for \(n=k+1\), whenever it is true for \(n=k\). is that sufficient to claim that the given statement is true for all integers : \(n=1,2,3,\ldots\) ?

  19. anonymous
    • one year ago
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    no?

  20. ganeshie8
    • one year ago
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    why no ?

  21. anonymous
    • one year ago
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    honestly idk

  22. ganeshie8
    • one year ago
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    if you know that the statement is true for \(n=k+1\), whenever it is true for \(n=k\). that essentially means the statement is true for ALL positive integers if you could establish it for \(n=1\)

  23. ganeshie8
    • one year ago
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    because, once you know that it is true for \(n=1\), it follows that it must be true for \(n=1+1=2\)

  24. ganeshie8
    • one year ago
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    after you know that it is true for \(n=2\), it follows that it must be true for \(n=2+1=3\)

  25. ganeshie8
    • one year ago
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    after you know that it is true for \(n=3\), it follows that it must be true for \(n=3+1=4\)

  26. ganeshie8
    • one year ago
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    see the pattern ? it follows that it must be true for ALL integers greater than 1

  27. anonymous
    • one year ago
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    ok i understand that

  28. ganeshie8
    • one year ago
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    So all we need to do is to establish two things : 1) the statement is true for \(n=1\) 2) the statement is true for \(n=k+1\), whenever it is true for \(n=k\)

  29. ganeshie8
    • one year ago
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    then by the previous induction patter, it will follow that the statement is true for ALL integers greater than 1

  30. ganeshie8
    • one year ago
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    You have already proven that the statement is true for \(n=1\) so lets look at the second step

  31. anonymous
    • one year ago
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    ok so what im doing is plugging in k+1?

  32. ganeshie8
    • one year ago
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    Suppose the statement is true \(n=k\), then we have : \[8 + 16 + 24 + ... + 8k = 4k(k + 1)\]

  33. ganeshie8
    • one year ago
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    add \(\color{red}{8(k+1)} \) both sides and get : \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4k(k + 1)\color{red}{+ 8(k+1)} \]

  34. ganeshie8
    • one year ago
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    factoring out 4(k+1) on right hand side, do we get : \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \] ?

  35. anonymous
    • one year ago
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    i'm sorry but all those parenthesis kinda confuses me.... O.O

  36. ganeshie8
    • one year ago
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    thats okay, take ur time

  37. anonymous
    • one year ago
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    i'm sorry but can you type it a bit more differently?.... is this what you're saying? 8+16+24+...+8k+8(k+1) = 4(k+1)(k+2)

  38. ganeshie8
    • one year ago
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    Yes, is there any issue with latex ?

  39. ganeshie8
    • one year ago
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    is it getting displayed like below on ur side ? https://i.gyazo.com/70406dea3fd312ded38a25e3f30da15b.png

  40. anonymous
    • one year ago
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    its coming out like this: \[8 + 16 + 24 + ... + 8k \color{red}{+ 8(k+1)} = 4(k + 1)(k+2) \]

  41. ganeshie8
    • one year ago
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    take a screenshot and attach, browser messes up the latex sometimes

  42. anonymous
    • one year ago
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    how do i screenshot it?

  43. ganeshie8
    • one year ago
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    https://gyazo.com/

  44. anonymous
    • one year ago
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    https://gyazo.com/e2eb27136301fb6cfe981d523de764f0

  45. ganeshie8
    • one year ago
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    Ahh that looks horrible, I'll give you the screenshot of complete proof, one sec

  46. ganeshie8
    • one year ago
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    Here is the complete proof https://i.gyazo.com/37b4e3e50a893db312121d7ed5de0944.png

  47. ganeshie8
    • one year ago
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    let me knw if something doesn't make sense

  48. anonymous
    • one year ago
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    oh ok i see.

  49. anonymous
    • one year ago
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    i understand it. thank you alot.

  50. ganeshie8
    • one year ago
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    yw

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