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anonymous

  • one year ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 1^2 + 4^2 + 7^2 + ... + (3n - 2)2 = quantity n times quantity six n squared minus three n minus one all divided by two

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  1. nuttyliaczar
    • one year ago
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    I suggest writing out that second half of the equation in terms like the first. ...=n(6n^2-3n-1)/2

  2. anonymous
    • one year ago
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    I know the statement for n=1 is true but I don't know how to find if k+1 is true

  3. nuttyliaczar
    • one year ago
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    Just check the equation again using the next integer, which would be 2

  4. nuttyliaczar
    • one year ago
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    Oh true true I forgot how you're supposed to prove by induction. Yes @Concentrationalizing is right to substitute all the n's with n+1 for the next part

  5. anonymous
    • one year ago
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    At that point what would I do next though?

  6. anonymous
    • one year ago
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    Okay, so we want to show this: \[1^{2} + 4^{2} + 7^{2} + ... + (3k-2)^{2} + [3(k+1)-2]^{2} = \frac{ (k+1)[6(k+1)^{2} - 3(k+1) - 1] }{ 2 }\]

  7. anonymous
    • one year ago
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    Sorry that it got cut off, its just a -1 in the numerator that got cut off.

  8. anonymous
    • one year ago
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    So by the induction hypothesis, we assumed this to be true: \[1^{2} + 4^{2} + 7^{2} + ... + (3k-2)^{2} = \frac{ k(6k^{2}-3k-1) }{ 2 }\]

  9. anonymous
    • one year ago
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    Thus we can do this substitution: \[\frac{ k(6k^{2} -3k-1) }{ 2 } + [3(k+1)-2]^{2} \]

  10. anonymous
    • one year ago
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    Now just to do the algebra and try to get it to look like k+1

  11. anonymous
    • one year ago
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    \[\frac{ k(6k^{2} - 3k-1) }{ 2 }+ [3(k+1)-2]^{2} = \frac{ k(6k^{2} - 3k - 1)+2(3k+1)^{2} }{ 2 }\]

  12. anonymous
    • one year ago
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    why do you add on the + [3(k+1)-2]^2

  13. anonymous
    • one year ago
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    Well, on the left side we have an infinite sum. Every consecutive term on the left must be added. When we do the k+1 term, its equivalent to replacing the k in (3k-2)^2 with k+1. But since its an infinite sum, that k+1 term must be added. Our original assumption though was that 1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k(6k^2-3k-1)/2 and you cannot simply add the [3(k+1)-2]^2 unles its added to both sides.

  14. anonymous
    • one year ago
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    Oh alright that makes sense

  15. anonymous
    • one year ago
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    Mhm. So the first thing I did was use our induction hypothesis to make the substitution on the left hand side. Now Im just trying to combine and simplify to make it look like the right.

  16. anonymous
    • one year ago
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    \[\frac{ k(6k^{2} - 3k - 1) }{ 2 } + [3(k+1)-2]^{2} = \frac{ 6k^{3} - 3k^{2} - k + 2(3k+1)^{2} }{ 2 }\] \[= \frac{ 6k^{3} - 3k^{2} - k + 2(9k^{2} + 6k + 1) }{ 2 }\] \[=\frac{ 6k^{3} - 3k^{2} - k + 18k^{2}+12k + 2}{ 2 } = \frac{ 6k^{3} +15k^{2}+11k+2 }{ 2 }\] Now here's where the tricky part comes in.

  17. anonymous
    • one year ago
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    Ill let you finish first

  18. anonymous
    • one year ago
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    so after the algebra my answer would be 1^2+4^2+7^2+... +(3k-2)^2+[3(k+1)-2]^2=(k+1)[6(k+1)^2-3(k+1)-1]/2

  19. anonymous
    • one year ago
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    Thats what youre hoping to get to. We need to manipulate the left hand side to make it look like the right, though. I have the left simplified down to what I have above.

  20. anonymous
    • one year ago
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    Where I have it right now is where it gets tricky, but here's what we can do. Now, notice the right hand side has a (k+1) multiple. Basically this means that k+1 is a factor of [6(k+1)^2 -3(k+1)-1], otherwise theres no way itd be factored out as it is. So what I want to do is take the numerator I have and divide it by k+1 and see what the remainder is

  21. anonymous
    • one year ago
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    So just a quick synthetic division: |dw:1433707447971:dw| So this means \[\frac{ 6k^{3} + 15k^{2} + 11k + 2 }{ 2 } = \frac{ (k+1)(6k^{2}+9k+2) }{ 2 }\]

  22. anonymous
    • one year ago
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    Bleh, lost what I was typing up, so gotta retype it, hang on

  23. anonymous
    • one year ago
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    no worries thank you so much!

  24. anonymous
    • one year ago
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    So, we need to get 6k^2 + 9k + 2 to look like 6(k+1)^2 -3(k+1) -1 If we were to expand that (don't actually expand this in your work, we don't want to mess with the right hand side, this is just to show you what we need to do) we would get 6k^2 + 12k + 6 -3k -3 -1 Conveniently, this is equivalent to what we have, so this is good. So let's just manipulate our left hand side to get this and then we can finish up \[\frac{ (k+1)(6k^{2}+9k +2) }{ 2 } = \frac{ (k+1)(6k^{2}+12k-3k + 6-3-1) }{ 2 }\] \[= \frac{ (k+1)(6k^{2}+12k+6 -3k-3 -1) }{ 2 } \] \[= \frac{ (k+1)[6(k^{2}+2k+1)-3(k+1)-1] }{ 2 }\] \[=\frac{ (k+1)[6(k+1)^{2} -3(k+1)-1] }{ 2 }\] So we got what we wanted. So you can conclude with something like: Therefore P(k+1) is true. Thus by the principal of mathematical induction, the statement P(n): 1^2 + 4^2 + 7^2 + ... + (3n-2)^2 = n(6n^2-3n-1)/n is true for all positive integers n. I feel like I butchered that a little, but honestly the tricky part is the algebra. I think the biggest trick to that was thinking to divide knowing that k+1 had to be a factor of what I was trying to simplify.

  25. anonymous
    • one year ago
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    Ive been trying to figure this out for a while now thanks a lot for your help!

  26. anonymous
    • one year ago
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    You're welcome

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