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Math Help - Need help simplifying

  1. #1
    Junior Member
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    Need help simplifying

    I was rushing on copying my notes and I do not know how this:

    5(35(3^n-1) - 29(2^n-1) ) - 6(35(3^n-2) - 29(2^n-2) )

    =
    (the 6 in the 2nd portion can be written as 2*3)
    5*35(3^n-1) - 5*29(2^n-1) - 2*35(3^n-1) + 3*29(2^n-1)

    this is where I do not know how I got this on my notes in the next step.

    = 3*35*(3^n-1) - 2*29(2^n-1)

    =35*(3^n) - 29(2^n)


    anyone know how I get = 3*35*(3^n-1) - 2*29(2^n-1), thanks in advance.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, ff4930!

    Quite a project . . .


    5\left[35\cdot3^{n-1} - 29\cdot2^{n-1}\right] - 6\left[35\cdot3^{n-2} - 29\cdot2^{n-2}\right)

    I do not know how I got this on my notes in the next step:
    . . 3\cdot35\cdot3^{n-1} - 2\cdot29\cdot2^{n-1} \;=\;35\cdot3^n - 29\cdot2^n

    5\left[35\cdot3^{n-1} - 29\cdot2^{n-1}\right] - 6\left[35\cdot3^{n-2} - 29\cdot2^{n-2}\right]

    . . = \;175\cdot3^{n-1} - 145\cdot2^{n-1} - 210\cdot3^{n-2} + 174\cdot2^{n-2}

    . . = \;\;\;\left[175\cdot3^{n-1} - 210\cdot3^{n-2}\right] \;- \;\left[145\cdot2^{n-1} - 174\cdot2^{n-2}\right]
    . . . . . . . . . . . . . \downarrow . . . . . . . . . . . . . . . . \downarrow
    . . = \;\left[175\cdot3^{n-1} - \overbrace{70\cdot3}\cdot3^{n-2}\right] - \left[145\cdot2^{n-1} - \overbrace{87\cdot2}\cdot2^{n-2}\right]

    . . = \;\left[175\cdot3^{n-1} - 70\cdot3^{n-1}\right] - \left[145\cdot2^{n-1} - 87\cdot2^{n-1}\right]

    . . = \;[175-70]\cdot3^{n-1} - [145-87]\cdot2^{n-1}

    . . = \;\;105\cdot3^{n-1} \;- \;58\cdot2^{n-1}
    . . . . . \downarrow . - . - . - . \downarrow
    . . = \;\overbrace{35\cdot3}\cdot3^{n-1} - \overbrace{29\cdot2}\cdot2^{n-1}

    . . = \quad35\cdot3^n\quad -\quad 29\cdot2^n

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  3. #3
    Junior Member
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    Thank you soooooo very much!!
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