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amorfide Group Title

So it is time to ask a question of my own, i do not need an entire solution, just what the next step is

  • 2 years ago
  • 2 years ago

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  1. amorfide Group Title
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    |dw:1350055553188:dw|

    • 2 years ago
  2. amorfide Group Title
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    find the values of A and B

    • 2 years ago
  3. amorfide Group Title
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    i did the first step which gives me

    • 2 years ago
  4. amorfide Group Title
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    |dw:1350055629792:dw|

    • 2 years ago
  5. amorfide Group Title
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    not sure what to do next

    • 2 years ago
  6. amorfide Group Title
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    @Callisto can you help?

    • 2 years ago
  7. seashell Group Title
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    @lgbasallote

    • 2 years ago
  8. lgbasallote Group Title
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    you tagged someone with an "i hate math" picture to a math question?

    • 2 years ago
  9. Callisto Group Title
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    Is it n(2n^2 + 3n -2) on the right side?

    • 2 years ago
  10. amorfide Group Title
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    yeah it is

    • 2 years ago
  11. Callisto Group Title
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    |dw:1350056495046:dw| Can you simplify it first?

    • 2 years ago
  12. amorfide Group Title
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    which side? left or right

    • 2 years ago
  13. Callisto Group Title
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    Both. There is a common factor.

    • 2 years ago
  14. amorfide Group Title
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    |dw:1350056560601:dw|

    • 2 years ago
  15. amorfide Group Title
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    that should be -2n on ther ight

    • 2 years ago
  16. amorfide Group Title
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    |dw:1350056701582:dw|

    • 2 years ago
  17. Callisto Group Title
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    Not really.\[\frac{A}{6}n(n+1)(2n+1) +bn= n(n^2 + 3n -2)\]\[n[\frac{A}{6}(n+1)(2n+1)+b] = n(n^2 + 3n -2)\] Cancel the common factor. and expand the left side.

    • 2 years ago
  18. amorfide Group Title
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    you could change b to 6b/6 and take out a factor of 1/6 right?

    • 2 years ago
  19. Callisto Group Title
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    You can, of course. But there is a common factor on both sides that you can cancel, what is it?

    • 2 years ago
  20. amorfide Group Title
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    get rid of n on both sides? if not then i dont know waht the common factor is on both sides

    • 2 years ago
  21. Callisto Group Title
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    Yes! So, now, you get \[\frac{A}{6}(n+1)(2n+1)+b = n^2 + 3n -2\]So, expand the terms on the left. What do you get?

    • 2 years ago
  22. amorfide Group Title
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    now i shall multiply out

    • 2 years ago
  23. Callisto Group Title
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    Yes.

    • 2 years ago
  24. amorfide Group Title
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    |dw:1350057038246:dw|

    • 2 years ago
  25. amorfide Group Title
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    forgot my A's

    • 2 years ago
  26. amorfide Group Title
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    |dw:1350057107864:dw|

    • 2 years ago
  27. Callisto Group Title
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    Not really..\[\frac{A}{6}(2n+1)(n+1) = \frac{A}{6}(2n^2+3n+1) = ...?\]

    • 2 years ago
  28. amorfide Group Title
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    oh and it is 2n² on the right

    • 2 years ago
  29. amorfide Group Title
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    |dw:1350057219369:dw|

    • 2 years ago
  30. amorfide Group Title
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    so that means to get 2n² A has to be 6

    • 2 years ago
  31. Callisto Group Title
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    I'm sorry but I have to go now... Basically, you need to expand the first term on the left, combine the like terms and compare the coefficient of x^2, x and constant terms on both sides. Then you can find A and B. As for your expansion, sorry again that it is wrong.

    • 2 years ago
  32. amorfide Group Title
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    okay thank you! i shall work on this thank you for your help

    • 2 years ago
  33. Callisto Group Title
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    a has to be 6 is correct, I supposed :|

    • 2 years ago
  34. Callisto Group Title
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    Once you find a, then compare the constant term on both sides. You should be able to get b. Good luck :)

    • 2 years ago
  35. amorfide Group Title
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    b is-3

    • 2 years ago
  36. amorfide Group Title
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    thank you :D

    • 2 years ago
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