Need help simplifying

**ff4930** · November 20th 2007, 03:08 PM

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.

**Soroban** · November 20th 2007, 05:00 PM

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:
. . $3\cdot35\cdot3^{n-1} - 2\cdot29\cdot2^{n-1} \;=\;35\cdot3^n - 29\cdot2^n$

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

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

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

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

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

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

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

**ff4930** · November 20th 2007, 05:04 PM

Thank you soooooo very much!!

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