Find sum of series (1/1*2)+(1/2*3)+(1/3*4)+(1/4*5)+ .............+(1/99*100)

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18 Jan 2013 Solve the Series Find sum of series (1/1*2)+(1/2*3)+(1/3*4)+(1/4*5)+ .............+(1/99*100): OPtion
1) 29/100
2) 39/100
3) 99/980
4) 59/900
5) 49/800
6) 99/100
7) 1/100
8) 69/100
9) 79/100
10) None of these; Correct Option: 6 Best Solution (10)

Dr.Dipin Singh 11 years AGO

(1/1*2)+(1/2*3)+(1/3*4)+(1/4*5)+ .............+(1/99*100) =

= [1/1 - 1/2] + [1/2 - 1/3] + [1/3 - 1/4] + ... + [1/98 - 1/99] + [1/99 - 1/100] = 1/1 - 1/100 = 99/100 ... option 6

Like? Yes (90) | No (31)

Vinay Sipani 11 years AGO

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

=> Sn = [1/1 - 1/2]+[1/2 - 1/3]+[1/3 - 1/4]+ ...
=> Sn = 1 - 1/(n+1)
=> Sn = n / (n + 1)

Here n = 99
=> Sn = 99 / 100

Like? Yes (43) | No (17)

Balakumaran N.A 11 years AGO

1/1*2+1/2*3+1/3*4+1/4*5+…+1/99*100
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/99-1/100
=1-1/100
=99/100

Like? Yes (18) | No (10)

Rohit Agarwal 11 years AGO

1/1*2 + 1/2*3 + 1/3*4 +........................1/99*100
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+................(1/99-1/100)
=1-1/100
=99/100

Like? Yes (13) | No (5)

Rats 11 years AGO

We can generalize the series as :
1/[(n)*(n+1)]=[1/n]-[1/(n+1)]
The series can be reconstructed as :
[1/1-1/2]+[1/2-1/3]+[1/3-1/4]+.....+[1/98-1/99]+[1/99-1/100]
We got
1/1-1/100(The remaining terms are cancelled out)
=99/100(Answer)

Like? Yes (15) | No (10)

chinna 11 years AGO