Telescoping Sum Of Fractions

Algebra Level 2

Evaluate

$\frac{ 1}{ 2(n-1) \times n } - \frac{ 1} { 2 n \times (n+1) }.$

\frac{ 1} { (n-1) \times (n+1) }

\frac{ 2n-1 } { (n-1) \times n \times (n+1) }

\frac{ 1} { (n-1) \times n \times (n+1) }

\frac{ 1}{n}

2 solutions

Chinmay Prabhakar
Apr 10, 2014

1/2 n {1/n-1 - 1/n+1}= 1/2n *{(n+1- (n-1))/(n-1) (n+1)} =1/2n* {2/(n-1) (n+1)} =1/n (n-1)*(n+1)

Vinayak Srivastava
May 10, 2020

\(\dfrac{1}{2(n-1)×n}-\dfrac{1}{2n×(n+1)}=

\dfrac{1}{(2n-2)×n}-\dfrac{1}{(2n^2+2n)}=

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

\dfrac{2n^2+2n}{(2n^2-2n)(2n^2+2n)}-\dfrac{2n^2-2n}{(2n^2+2n)(2n^2-2n)}=

\dfrac{2n^2+2n-2n^2+2n}{(2n^2-2n)(2n^2+2n)}=

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

\dfrac{4n}{(4n^4-4n^2)}=

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

\dfrac{4n}{(4n^4-4n^2)}=

\dfrac{1}{n(n^2-1)}=

\boxed{\dfrac{1}{n×(n+1)×(n-1)}}\)

Side note: Taking common denominator would reduce some of the arithmetic (and hence possibility of errors).

Calvin Lin Staff - 1 year, 1 month ago