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The problem asks to find the sum of the following series:

\[
\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \dots + \frac{1}{99 \times 103}
\]

This is a telescoping series where the general term is of the form:

\[
\frac{1}{(4n-1)(4n+3)}
\]

The series simplifies using partial fractions. For each term:

\[
\frac{1}{(a)(b)} = \frac{1}{a - b} \left(\frac{1}{a} - \frac{1}{b}\right)
\]

So, let's apply this to each term of the series and find the sum.

We can express each fraction as:

\[
\frac{1}{a(a+4)} = \frac{1}{4} \left( \frac{1}{a} - \frac{1}{a+4} \right)
\]

When summed, the intermediate terms cancel out, leaving only the first and last terms. The series effectively collapses to:

\[
\frac{1}{3} - \frac{1}{103}
\]

Now, let's compute this:

\[
\frac{1}{3} - \frac{1}{103} = \frac{103 - 3}{309} = \frac{100}{309}
\]

Thus, the sum of the series is:

\[
\boxed{\frac{100}{309}}
\]

Would you like more details on the method, or do you have any questions?

Here are 5 related questions to consider:

1. How do you find the sum of an infinite telescoping series?
2. Can you generalize this method for any product of consecutive integers?
3. What are partial fractions, and why are they useful in simplifying sums?
4. How does this telescoping series differ from a geometric series?
5. What applications do telescoping series have in other areas of mathematics?

**Tip:** When dealing with series, always check if terms can cancel out, as it simplifies the summation greatly!

Find the sum of 1/(3×7) + 1/(7×11) + 1/(11×15) + ... + 1/(99×103)

Solve the sum of the telescoping series from 1/(3×7) to 1/(99×103) using partial fractions. This problem involves key concepts in algebra and simplifies to a fraction. Suitable for high school students in Grades 11-12.

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