Reducing a fraction with exponents

Question

How do I reduce

$$ \frac{(k+1)^2}{k^2} $$

to simplest terms? My algebra is really rusty...

Thanks!

Answer 1

That fraction is already in simplest terms, since $k$ and $k+1$ don't share any common factors.

Answer 2

HINT:

Observe that if integer $d$ divides both $k+1$ and $k,$

$ d$ will divide $(k+1-k)=1\implies (k+1,k)=1$

Now, $(a^n,b^n)=(a,b)^n$

Answer 3

How about $$\frac{(k+1)^2}{k^2} = \left(\frac {k+1}{k}\right)^2$$ since $\;\dfrac{a^n}{b^n} = \left(\dfrac ab\right)^n$

That's about as simple as it gets.

Answer 4

$\frac{(k+1)^2}{k^2}$=$\frac{k^2+2k+1}{k^2}$=$1+\frac{2}{k}+\frac{1}{k^2}$