Math Doubts

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

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

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


Basic simplification

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.

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


Simplification of first two terms

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

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

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

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

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

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

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


Simplification of the remaining terms

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

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

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.

Latest Math Topics
Latest Math Problems
Email subscription
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more