Telescoping Series 4 – Generalizing the Pattern

Algebra Level 1

Using the pattern identified in the previous problems , find the values of $e$ and $f$ in terms of $n$ :

$\frac{1}{n(n+1)} = \frac{1}{e} - \frac{1}{f}$

e = n+1

and

f = n

e = n

and

f = n+1

e = n-1

and

f = n

e = n

and

f = n-1

6 solutions

Krishna Garg
Apr 9, 2014

here denominators are n and n+1 therefore e +n and f =n=1 Ans

K.K.GARG.India

yo yo, i solved it again

Armaan Das - 7 years, 1 month ago

1/e-1/f=1/n-1/n+1=(n+1-n)/n(n+)=1/n(n+1)...........so e=n and n+1is correct

Mukhtar Ahmad - 7 years, 2 months ago

Fahim Khan
Apr 29, 2014

This is the theory 1/n(n+1)= (1/n) - (1/n+1)

Sara Rahman
Apr 13, 2014

since we know that 1/1*2=1/1-1/2,from this relation i can conclude that e=n,f=n+a

Harshith Balasubramanian
Apr 10, 2014

\cfrac { 1 }{ n(n+1) } =\frac { 1 }{ e } -\frac { 1 }{ f }

Abdul Basit Idrees
Apr 10, 2014

The problem solve by this equation. {1/(n*(n+1))=(1/n)-(1/n+1)} and the denominators are; e=n and f=n+1

Abid Ali
Apr 10, 2014

denominators are n and n+1 therefore e +n and f =n=1

The pattern it self gives the solution.no need of too much maths knowledge.

Dommeti NagavenkataSatyanarayana - 7 years, 1 month ago