$n$ is a literal number and an algebraic expression is formed to represent a quantity in mathematical form.

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

01

A basic simplification is required to simply the sum of the three terms.

$=\,\,\,$ $\dfrac{1}{\Bigg(\dfrac{n-1}{2}\Bigg)}$ $+$ $\dfrac{1}{\Bigg(\dfrac{n+1}{2}\Bigg)}$ $-$ $\dfrac{1}{\dfrac{n(n-1)(n+1)}{2 \times 2}}$

$=\,\,\,$ $\dfrac{1}{\Bigg(\dfrac{n-1}{2}\Bigg)}$ $+$ $\dfrac{1}{\Bigg(\dfrac{n+1}{2}\Bigg)}$ $-$ $\dfrac{1}{\dfrac{n(n-1)(n+1)}{4}}$

According to reciprocal rule, each term can be written as follows.

$=\,\,\,$ $\dfrac{2}{n-1}$ $+$ $\dfrac{2}{n+1}$ $-$ $\dfrac{4}{n(n-1)(n+1)}$

02

The technique in this problem is to solve the first two terms firstly. Take $2$ common from both terms and simplify them.

$=\,\,\,$ $2\Bigg[\dfrac{1}{n-1} + \dfrac{1}{n+1}\Bigg]$ $-$ $\dfrac{4}{n(n-1)(n+1)}$

$=\,\,\,$ $2\Bigg[\dfrac{n+1+n-1}{(n-1)(n+1)}\Bigg]$ $-$ $\dfrac{4}{n(n-1)(n+1)}$

$=\,\,\,$ $\require{cancel} 2\Bigg[\dfrac{n+\cancel{1}+n-\cancel{1}}{(n-1)(n+1)}\Bigg]$ $-$ $\dfrac{4}{n(n-1)(n+1)}$

$=\,\,\,$ $2\Bigg[\dfrac{2n}{(n-1)(n+1)}\Bigg]$ $-$ $\dfrac{4}{n(n-1)(n+1)}$

$=\,\,\,$ $\dfrac{2 \times 2n}{(n-1)(n+1)}$ $-$ $\dfrac{4}{n(n-1)(n+1)}$

$=\,\,\,$ $\dfrac{4n}{(n-1)(n+1)}$ $-$ $\dfrac{4}{n(n-1)(n+1)}$

03

$=\,\,\,$ $\dfrac{4n \times n -4}{n(n-1)(n+1)}$

$=\,\,\,$ $\dfrac{4n^2-4}{n(n-1)(n+1)}$

Take the number $4$ common from both terms in the numerator.

$=\,\,\,$ $\dfrac{4(n^2-1)}{n(n-1)(n+1)}$

$=\,\,\,$ $\dfrac{4(n^2-1^2)}{n(n-1)(n+1)}$

$=\,\,\,$ $\dfrac{4(n-1)(n+1)}{n(n-1)(n+1)}$

$=\,\,\,$ $\require{cancel} \dfrac{4\cancel{(n-1)}\cancel{(n+1)}}{n\cancel{(n-1)}\cancel{(n+1)}}$

$=\,\,\,$ $\dfrac{4}{n}$

It is the required solution after simplifying the algebraic expression in mathematics.

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