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How did this conversion took place please help?

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    6(2)^(n-2) +1

    then summation of above equation will be
    6(2)^(n-1) + n + 2

    can you please tell me how this equaton came from the first equation?

    3. The attempt at a solution

    the later formula is correct summation of the former

    because it may look that the summation is
    6(2)^(n-1) +n - 3

    but by putting values of n-
    6(2)^(n-1) +n - 3 is incorrect

    please help as how the topmost formula gives the second formula
     
    Last edited: Feb 9, 2012
  2. jcsd
  3. Feb 9, 2012 #2

    Mark44

    Staff: Mentor

    It doesn't. First off, these aren't equations - they are expressions. An equation is a statement that two expressions are equal, and always has an = symbol in it.

    Starting from your first expression,
    6[itex]\cdot[/itex]2n-2 + 1 = 6[itex]\cdot[/itex]2n-1[itex]\cdot[/itex](1/2) + 1 = 3[itex]\cdot[/itex]2n-1 + 1

    To show that your two expressions above aren't identically equal, let n = 2. Then 6[itex]\cdot[/itex]2n-2 + 1 = 6 * 1 + 1 = 7.

    And 6[itex]\cdot[/itex]2n-1 + n + 2 = 6* 2 + 2 + 2 = 16.

     
  4. Feb 9, 2012 #3

    eumyang

    User Avatar
    Homework Helper

    I'm not sure what these expressions (and these are expressions, not equation) are. Are there any symbols missing? Can you attach an image, or retype using LaTeX?
     
  5. Feb 10, 2012 #4
  6. Feb 10, 2012 #5

    eumyang

    User Avatar
    Homework Helper

    I get a "page is unavailable" error.
     
  7. Feb 11, 2012 #6
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