IIT JEE , AIEEE Forum - Mathematics , Algebra

vivek_aero

How to find the summation of 1/n(n+1)

iberis22

1/n(n+1)

= (1 + n - n)/ n(n+1) [Adding and subtracting n in the numerator]

= (n+1)/n(n+1) - n/n(n+1)

= 1/n - 1/(n+1)

on taking summation, n=1 to r

= [1/1 + 1/2 + 1/3 + 1/4 + 1/5 .... + 1/(r-1) + 1/r ] - [1/2 + 1/3 + 1/4 + 1/5 .... + 1/r + 1/(r+1)]

= 1 - 1/(r+1)

= r/(r+1)

ankur.kkhurana

in general divide the numerator in to factors in denominator and divide about them after it solve summation

magiclko

such questions are done by the method of partial fractions..

1/ [n(n+1)] can be written as (1/n) - (1/(n+1))

that is T_n = (1/n) - (1/(n+1))

thrfore

T₁ = 1 - 1/2

T₂ = 1/2 - 1/3

T₃ = 1/3 - 1/4

T₄ = 1/4 - 1/5

.

T_n-1 = (1/n-1) - (1/n)

T_n = (1/n) - (1/(n+1))

adding all we have S = 1 - 1/(n+1)

= n/n+1