Equating two functions

[ericbakuladavis]

Hi,

In writing a rule for a sequence, I came up with two functions:

f(n) = 2^n + ((2^(n-1)) / 10)

and

f(n) = 2.1 (2^(n-1))

They both seem to describe the sequence. Is it true that...

2^n + ((2^(n-1)) / 10) = 2.1 (2^(n-1)) ?

If so, how can this be proven? [EDIT: I was having trouble imagining how to even begin rearranging either function to equal the other. I thought that maybe the n's on the left side need to be combined but I couldn't figure out how.]

Thanks,

Eric

 

[JeffM]

Hi,

In writing a rule for a sequence, I came up with two functions:

f(n) = 2^n + ((2^(n-1)) / 10)

and

f(n) = 2.1 (2^(n-1))

They both seem to describe the sequence. Is it true that...

2^n + ((2^(n-1)) / 10) = 2.1 (2^(n-1)) ?

If so, how can this be proven?

Thanks,

Eric

\(\displaystyle 2^n + \dfrac{2^{(n - 1)}}{10} = \dfrac{10 * 2^n}{10} + \dfrac{2^{(n - 1)}}{10} \implies\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac{(10 * 2^n) + 2^{(n - 1)}}{10}\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac{(10 * 2 * 2^{(n - 1)}) + 2^{(n - 1)}}{10}\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac{20 * 2^{(n - 1)} + 2^{(n - 1)}}{10}\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac{(20 + 1) * 2^{(n - 1)}}{10}\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \dfrac{21 * 2^{(n - 1)}}{10}\)

\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.1 * 2^{(n - 1)}\)

 

[ericbakuladavis]

Brilliant! Thank you. The step that really impressed me was going from 2^n to 2 * 2^(n-1). I never learned that one in class...

 

[JeffM]

Brilliant! Thank you. The step that really impressed me was going from 2^n to 2 * 2^(n-1). I never learned that one in class...

You're welcome. Glad to have been of help.

You hear that denis and subhotosh. I am brilliant: no more sending me to the corner. Lesser minds such as yours simply do not grasp my capacities. I used to tell that to my employer (back when I as still employed, that is).