To solve the problem of summing fractions of the form (\frac{1}{k(k+1)}) (a common telescoping series), follow these steps:

Step 1: Partial Fraction Decomposition

Rewrite each term using partial fractions:
[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} ]
This is derived by solving (\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}), leading to (A=1) and (B=-1).

Step 2: Telescoping Sum

Sum the terms from (k=1) to (n):
[ \sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+1}\right) = \left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \dots + \left(\frac{1}{n}-\frac{1}{n+1}\right) ]

Most terms cancel out (telescope), leaving:
[ 1 - \frac{1}{n+1} = \frac{n}{n+1} ]

Example (if the problem asks for sum up to 100):

[ \frac{100}{101} ]

General Answer: (\boxed{\frac{n}{n+1}}) (replace (n) with the upper limit in the problem).

If the problem specifies an upper limit (e.g., 100), the answer is (\boxed{\frac{100}{101}}).

Assuming the common case of upper limit 100, the final answer is:
(\boxed{\dfrac{100}{101}})

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