# How to prove this equation using mathematical induction? I need to prove the equation 1

How to prove this equation using mathematical induction?
I need to prove the equation
$\frac{1}{3}=\frac{1+3+5+\cdots +\left(2n-1\right)}{\left(2n+1\right)+\left(2n+3\right)+\cdots +\left(2n+\left(2n-1\right)\right)}$ 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 $n=k$
$\frac{1}{3}=\frac{1+3+5+\cdots +\left(2k-1\right)}{\left(2k+1\right)+\left(2k+3\right)+\cdots +\left(2k+\left(2k-1\right)\right)}$
- For $n=k+1$
$\frac{1}{3}=\frac{1+3+5+\cdots +\left(2k-1\right)+\left(2k+1-1\right)}{\left(2k+1\right)+\left(2k+3\right)+\cdots +\left(2k+\left(2k-1\right)\right)}$
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Punktatsp
Step 1
You did some mistake in the expression for the case $n=k+1$ in the induction step. Morover it is convenient proceed as follows using that $\frac{A}{B}=\frac{1}{3}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3A=B$.
We need to show by induction that that for any $n\ge 1$.
$\frac{1}{3}=\sum _{i=1}^{n}\frac{1+3+5+\cdots +\left(2i-1\right)}{\left(2i+1\right)+\left(2i+3\right)+\cdots +\left(2i+\left(2i-1\right)\right)}$
- base case: $n=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{1}{3}=\frac{1}{2+1}$
- induction step we assume
$\frac{1}{3}=\sum _{i=1}^{n}\frac{1+3+5+\cdots +\left(2i-1\right)}{\left(2i+1\right)+\left(2i+3\right)+\cdots +\left(2i+\left(2i-1\right)\right)}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3\sum _{i=1}^{n}\left(1+3+5+\cdots +\left(2i-1\right)\right)=\sum _{i=1}^{n}\left(2i+1\right)+\left(2i+3\right)+\cdots +\left(2i+\left(2i-1\right)\right)$
Step 2
Then $3\sum _{i=1}^{n+1}\left(1+3+5+\cdots +\left(2i-1\right)\right)=\sum _{i=1}^{n+1}\left(2i+1\right)+\left(2i+3\right)+\cdots +\left(2i+\left(2i-1\right)\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3\left(2\left(n+1\right)-1\right)=2n+2\left(n+1\right)+\left(2\left(n+1\right)-1\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3\left(2n+1\right)=2n+2n+2+\left(2n+1\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}6n+3=4n+2+\left(2n+1\right)$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}6n+3=6n+3$

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