Simplifying (1+x)^(k-1) / (1+x)^(2k) to 1/(1+x)^(k+1)

[sony]

How can you go from:

(1+x)^(k-1) / (1+x)^(2k)
to:
1/ (1+x)^(k+1)

?

Thanks!

 

[VietDao29]

You can use:
$$\frac{a ^ \alpha}{a ^ \beta} = a ^ {\alpha - \beta}$$
And $$a ^ {- \alpha} = a ^ {0 - \alpha} = \frac{a ^ 0}{a ^ \alpha} = \frac{1}{a ^ \alpha}$$ to solve the problem. So:
$$\frac{(1 + x) ^ {k - 1}}{(1 + x) ^ {2k}} = (1 + x) ^ {k - 1 - 2k} = ...$$
Can you go from here?
Viet Dao,

 

[sony]

= (1+x)^(-k-1)
=1 / (1+x)^(k+1)

Right?

 

[VietDao29]

Yup. That's correct.
Viet Dao,

 

[sony]

Ok, thanks :)