Very quick question about powers

[oshea.emma]

I'm working with some difference equations and there's something that's confusing me that involves a power.

I'm told that this is the solution but I don't fully understand it.

Is there anyone out there that can help?

How does this work?

Would it not be correct to say that it should be n and not n-1 because the parts are being multiplied?

?

 

[TchrWill]

How does this work?

Would it not be correct to say that it should be n+1 and not n-1 because the parts are being multiplied?

2 + 6(1/6)^n = 2 + (36/6)^1(1/6)^n = 2 + 36(1/6)^(n + 1)]

 

[oshea.emma]

To the best of my knowledge the answer outlined is correct.

?

So it should be n-1...or are you saying that this is wrong?

 

[galactus]

Yes, it is correct.

\(\displaystyle \L\\\left(\frac{1}{6}\right)^{n-1}=\frac{1}{6^{n}}\cdot\underbrace{\frac{1}{6^{-1}}}_{\text{this is 6}}=6\cdot\frac{1}{{6}^{n}}=6\left(\frac{1}{6}\right)^{n}\)

 

[TchrWill]

To the best of my knowledge the answer outlined is correct.?

So it should be n-1...or are you saying that this is wrong?

a^m x a^n = a^(m+n)

(1/6)^1 x (1/6)n = (1/6)^(n + 1)

Example:
2 + 6(1/6)^4 = 2.00462963

2 + (36)(1/6)^5 = 2.00462963

 

[oshea.emma]

I'm a little confused.

Which answer is correct?

galactus

\(\displaystyle \L\\\left(\frac{1}{6}\right)^{n-1}=\frac{1}{6^{n}}\cdot\underbrace{\frac{1}{6^{-1}}}_{\text{this is 6}}\)

Why is this correct?

When you add them wouldn't the answer become...\(\displaystyle \frac{1}{6^{n-1}}\\)

And then would the sign change when you invert?

Maybe you could explain?

Thanks...

 

[TchrWill]

To the best of my knowledge the answer outlined is correct.?

So it should be n-1...or are you saying that this is wrong?

Galactus is correct also.

2 + 36(1/6)^(n + 1) is equivalent to 2 + (1/6)^(n - 1)

 

[Mrspi]

(1/6)<SUP>n-1</SUP> = (1/6)<SUP>n</SUP>*(1/6)<SUP>-1</SUP>

But (1/6)<SUP>-1</SUP> = 6<SUP>1</SUP> by the definition of a negative exponent.

So,
(1/6)<SUP>n</SUP> * (1/6)<SUP>-1</SUP> = (1/6)<SUP>n</SUP> * 6, or 6*(1/6)<SUP>n</SUP>

Thus,
2 + 6*(1/6)<SUP>n</SUP> can be written as 2 + (1/6)<SUP>n - 1</SUP>

 

[oshea.emma]

Thank you, that makes sense now! Much appreciated.