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The sum of n from 1 to 99 for all n 1/ (n(n+1))
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How do you find the sum of.. 1/1.2+1/2.3+1/3.4.... +1/99.100 ???? plss help me..
try factorization
experimentX plss help me.. i dont know to how find the sum..
\[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]Now use telescoping ...
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
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what is the solution?
\[ \frac 11 - \frac12 + \frac 12 - \frac 13 +\frac 13 + ... + \frac 1{99} - \frac 1 {100} \]
99/100 is the solution
Thank you ajaykharabe and experimentX....
Thanks for helping me.. :)
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