Summation of fractions with odd denominators to prove by induction

Question

$$ \frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\dots+\frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1} $$

As you can imagine I am stuck in third step in $k+1$. Hope you can help. 

Answer 1

First, show that this is true for $n=1$:

$\sum\limits_{k=1}^{1}\dfrac{1}{(2k-1)(2k+1)}=\dfrac{1}{2+1}$

Second, assume that this is true for $n$:

$\sum\limits_{k=1}^{n}\dfrac{1}{(2k-1)(2k+1)}=\dfrac{n}{2n+1}$

Third, prove that this is true for $n+1$:

$\sum\limits_{k=1}^{n+1}\dfrac{1}{(2k-1)(2k+1)}=$

$\color\red{\sum\limits_{k=1}^{n}\dfrac{1}{(2k-1)(2k+1)}}+\dfrac{1}{(2(n+1)-1)(2(n+1)+1)}=$

$\color\red{\dfrac{n}{2n+1}}+\dfrac{1}{(2(n+1)-1)(2(n+1)+1)}=$

$\dfrac{n+1}{2(n+1)+1}$

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Please note that the assumption is used only in the part marked red.