1. ## Factoring 2^{^{n-3}}-2^{^{n-2}}

Hi,

So right now I'm studying for an exam and I'm doing a problem on successions.

ln+1 - ln

is positive or negative.

ln = 1/(2^(n-3))

I was able to solve most of it, but then I got stuck; so I looked up a solution in symbolab:

https://www.symbolab.com/solver/polynomial-equation-calculator/%5Cfrac%7B1%7D%7B2%5E%7B%5E%7B%5Cleft(n%2B1%5Crigh t)-3%7D%7D%7D-%5Cfrac%7B1%7D%7B2%5E%7B%5E%7Bn-3%7D%7D%7D

2^(^(n-3))-2^(^{n-2))

where the website factors out the common term 2^(-3)*2^n of

2^(^(n-3))-2^(^{n-2))

to get

-2^(n-3)

Can someone help me understand this factoring? I'm confused by how the site got to this final answer. How does it factor out
2^(-3)*2^n ?

If you can help me understand this I would really appreciate it.

P.S. Sorry if the formatting is horrible. I'm still figuring out how to use the site

2. Originally Posted by skatinima
Hi,
ln+1 - ln

is positive or negative.

ln = 1/(2^(n-3))

2^(^(n-3))-2^(^{n-2))

where the website factors out the common term 2^(-3)*2^n of

2^(^(n-3))-2^(^{n-2))

to get

-2^(n-3)

Typically, just piece it together. At first, we're just finding a common denominator.

$\dfrac{1}{2^{n-3}} - \dfrac{1}{2^{n-2}} = \dfrac{1}{2^{n-3}} - \dfrac{2}{2^{n-3}}$

Do you see that this is so? Looks like we're almost done and we haven't done any factoring.

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