[administration]

Q. 1/(1*2*3) +1/(3*4*5) + 1/(5*6*7) + ……..

Find the  sum ?

[Wonder Boy]

To solve  this  type of problem, where reciprocals are involved we will use the concept of partial fraction.

so here we can see each term is of the form

¹/[_{n.(n+1).(n+2)]}

Let ¹/_{[n.(n+1).(n+2)]}  =  ^A/_n + ^B/_(n+1) + ^C/_(n+2)

–> ¹/_{[n.(n+1).(n+2)]} = ^{[ A(n+1)(n+2) + B(n)(n+2) +C(n)(n+1)]}/_{[n.(n+1).(n+2)]} 

–> 1 = [ A(n+1)(n+2) + B(n)(n+2) +C(n)(n+1)]

putting n=-1, we get

B = -1

putting n = -2 , we get

C = 1/2

putting n= 0, we get

A= 1/2

So , ¹/_{[n.(n+1).(n+2)]} = ¹/_2n –¹/_(n+1) + ¹/_2(n+2)

1/(1*2*3) +1/(3*4*5) + 1/(5*6*7) + …….. = 1/2.1-1/2+1/2.3   +  1/2.3-1/4+1/2.5  +  1/2.5-1/6+1/2.7 +……

= 1/2(1+1/3 + 1/5+1/7 + …..) – (1/2 +1/4 +1/6 +…… ) + 1/2(1/3 +1/5 +1/7 +…..)

= 1/2(1+1/3 + 1/5+1/7 + …..) – (1/2 +1/4 +1/6 +…… ) -1/2

= (1-1/2+1/3-1/4+ ….)  -1/2

= 0.69314 -0.5 = 0.19314

 

 

 

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