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anonymous

  • 5 years ago

Hi, I need to prove Sn= 6n^2-6n+1. How can I d othis?

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  1. anonymous
    • 5 years ago
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    \[6n ^{2}-6n+1\] HOW CAN I PROVE THIS?

  2. anonymous
    • 5 years ago
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    Need more info, what subject or course are you doing?

  3. anonymous
    • 5 years ago
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    Just Math, an IB course. I don't know how to prove this... at all, any questions to lead me towards something productive?

  4. anonymous
    • 5 years ago
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    6n^2-6n+1 = 0 is that what yu are trying to prove?

  5. anonymous
    • 5 years ago
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    not enough info o_o

  6. anonymous
    • 5 years ago
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    What is Sn?

  7. anonymous
    • 5 years ago
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    f(1) = 1 f(2) = 13 f(3) = 37 f(4) = 73 f(5) = 121 f(6) = 181

  8. anonymous
    • 5 years ago
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    It's actually, f(n) = 6n^2 - 6 + 1

  9. anonymous
    • 5 years ago
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    wait, ANOTHER TYPO. f(n) = 6n^2 - 6n + 1 a quadratic equation.

  10. anonymous
    • 5 years ago
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    Oh I see.

  11. anonymous
    • 5 years ago
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    haha now it's solvable ^^

  12. anonymous
    • 5 years ago
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    so how can I do this?

  13. anonymous
    • 5 years ago
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    get any point yu have and put on your equantion: f(5) = 121 121 = 6(5)^2 - 6(5) + 1 121 = 150 - 30 + 1 121 = 121 (True) is that right?

  14. anonymous
    • 5 years ago
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    But that's just checking, I need to know how to prove the equation without knowing the formula already..?

  15. amistre64
    • 5 years ago
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    either factor it or complete the square

  16. anonymous
    • 5 years ago
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    i think she has to find the formula with those points she has

  17. anonymous
    • 5 years ago
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    1st term=12+1 2nd term=13+2(12) 3rd term=37+3(12) 4th term=73+4(12) . . nth term=(n-1)th term+n(12)

  18. anonymous
    • 5 years ago
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    Oh sorry, first term is 1. So, the formula should be: \[a_n=a_{n-1}+12(n-1)\]

  19. anonymous
    • 5 years ago
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    and then.. ? what do I do from there to get 6n^2 - 6n + 1

  20. anonymous
    • 5 years ago
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    Look above at the example of Mathmind, that is what you do for each term. You just dismissed it as that's just checking. That is what you are supposed to do, check to see if it is true or not.

  21. anonymous
    • 5 years ago
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    so how would I prove the equation if I don't know what the equation is to begin with...

  22. anonymous
    • 5 years ago
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    You were given the info. You gave us the info. I think you are a little lost, but that is OK. When they say f(1)=1. They are saying that 6(1)^2-6(1)+1=1 I this true or not?

  23. anonymous
    • 5 years ago
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    Yeah I know that's true. But I'm not proving that the points match, I'm using the points to find an equation, say that I don't know what the equation is.

  24. anonymous
    • 5 years ago
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    f(1)=1 f(2)=1+12 f(3)=1+12+2(12)=1+12(1+2) f(4)=1+12+2(12)+3(12)=1+12(1+2+3) f(n)=1+12(1+2+3+..(n-1))

  25. anonymous
    • 5 years ago
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    Now we are lost. You said you wanted to prove something. You didn't say you want to find an equation. We apologize.

  26. anonymous
    • 5 years ago
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    There is a formula for the summation (1+2+3+..+(n-1)), that's: \[1+2+3+...+(n-1)={1 \over 2}n(n-1)\] Plug that in the previous equation, you get: \[f(n)=1+12({1 \over 2}n(n-1))=1+6(n(n-1))=1+6n^2-6n\]

  27. anonymous
    • 5 years ago
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    I hope that helps.

  28. anonymous
    • 5 years ago
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    THANK YOU SO MUCH !

  29. anonymous
    • 5 years ago
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    Can you help me explain this? I'm not sure how I would explain this

  30. anonymous
    • 5 years ago
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    The steps are all clear. Try to read them again, the last two comments before "I hope that hels".

  31. anonymous
    • 5 years ago
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    If there is something you don't understand, let me know.

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