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
Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved