To solve the problem involving the sum of reciprocals of products of consecutive integers, we use the telescoping series method. Here's the step-by-step breakdown:

Key Insight: Partial Fraction Decomposition

For any integer (k), the term (\frac{1}{k(k+1)}) can be split into:
[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} ]

Sum Calculation

Consider the sum (S = \sum_{k=1}^n \frac{1}{k(k+1)}):
[ S = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right) ]

All intermediate terms cancel out (telescoping effect):
[ S = 1 - \frac{1}{n+1} = \frac{n}{n+1} ]

Example Result

If the sum is up to (n=99) (e.g., last term (\frac{1}{99×100})), the result is:
[ \frac{99}{100} ]

Answer: (\boxed{\frac{n}{n+1}}) (replace (n) with the upper limit of the series; for common cases like (n=99), it's (\boxed{\frac{99}{100}})).

Assuming the problem's upper limit is 99, the final answer is:
(\boxed{\frac{99}{100}}) (adjust based on the actual (n) in the image, but this is a typical result).

(\boxed{\dfrac{99}{100}})

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