How to prove this equation using mathematical induction?

I need to prove the equation

$\frac{1}{3}=\frac{1+3+5+\cdots +(2n-1)}{(2n+1)+(2n+3)+\cdots +{\textstyle (}2n+(2n-1){\textstyle )}}$ using mathematical induction.

I tried solving this but I got stuck. I would be very thankful is someone could help me. Or maybe give me a references or hint.

My Attempt:

- For $n=1$

$\frac{1}{3}=\frac{1}{2n+1}$

$\frac{1}{3}=\frac{1}{2+1}$

$\frac{1}{3}=\frac{1}{3}$ which is true.

- For $n=k$

$\frac{1}{3}=\frac{1+3+5+\cdots +(2k-1)}{(2k+1)+(2k+3)+\cdots +{\textstyle (}2k+(2k-1){\textstyle )}}$

- For $n=k+1$

$\frac{1}{3}=\frac{1+3+5+\cdots +(2k-1)+(2k+1-1)}{(2k+1)+(2k+3)+\cdots +{\textstyle (}2k+(2k-1){\textstyle )}}$

I need to prove the equation

$\frac{1}{3}=\frac{1+3+5+\cdots +(2n-1)}{(2n+1)+(2n+3)+\cdots +{\textstyle (}2n+(2n-1){\textstyle )}}$ using mathematical induction.

I tried solving this but I got stuck. I would be very thankful is someone could help me. Or maybe give me a references or hint.

My Attempt:

- For $n=1$

$\frac{1}{3}=\frac{1}{2n+1}$

$\frac{1}{3}=\frac{1}{2+1}$

$\frac{1}{3}=\frac{1}{3}$ which is true.

- For $n=k$

$\frac{1}{3}=\frac{1+3+5+\cdots +(2k-1)}{(2k+1)+(2k+3)+\cdots +{\textstyle (}2k+(2k-1){\textstyle )}}$

- For $n=k+1$

$\frac{1}{3}=\frac{1+3+5+\cdots +(2k-1)+(2k+1-1)}{(2k+1)+(2k+3)+\cdots +{\textstyle (}2k+(2k-1){\textstyle )}}$