IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

deep01

evaluate

_{[n ]}

_{[ ]} [1/(1.2)+1/(2.3)+1/(2.4)+-------------------1/(n(n+1))

pls also epxlain ur answer

kusum

hi ,

the general term is 1/r(r+1) ,

which if resolved in partial fractions

1/r(r+1)=1/r - 1/(r+1)

now u can understand if we sum ,

1-1/2 +1/2 -1/3 ............-1/n +1/n - 1/n+1

=1- 1/(n+1)

=1 (as n tending to infinity 1/(n+1) will tend to zero.)

i hope i am clear .

iitkgp_bipin

1/(1.2) = (1 - 1/2)

1/(2.3) = (1/2 - 1/3)

1/(3.4) = (1/3 - 1/4)

...........

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

If we add all these terms 1/2 , 1/3 , ....... , 1/n gets cancelled.

Sum = (1 - 1/n+1)

As n tends to infinity the sum becomes 1

So answer is 1

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