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Mathematics 13 Online
OpenStudy (anonymous):

The sum of n from 1 to 99 for all n 1/ (n(n+1))

OpenStudy (anonymous):

How do you find the sum of.. 1/1.2+1/2.3+1/3.4.... +1/99.100 ???? plss help me..

OpenStudy (experimentx):

try factorization

OpenStudy (anonymous):

experimentX plss help me.. i dont know to how find the sum..

OpenStudy (experimentx):

\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]Now use telescoping ...

OpenStudy (anonymous):

Just try some first (my way) like (1/1.2)+(1/2.3)=2/3 so it becomes denominator of last number making first numerator so ans is 99/100

OpenStudy (anonymous):

what is the solution?

OpenStudy (experimentx):

\[ \frac 11 - \frac12 + \frac 12 - \frac 13 +\frac 13 + ... + \frac 1{99} - \frac 1 {100} \]

OpenStudy (anonymous):

99/100 is the solution

OpenStudy (anonymous):

Thank you ajaykharabe and experimentX....

OpenStudy (anonymous):

Thanks for helping me.. :)

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