To find the sum of the series 

\[
A = \frac{1}{3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \cdots + \frac{1}{97 \times 99}
\]

we can observe that each term in the series has the form:

\[
\frac{1}{n \times (n+2)}
\]

where \( n = 3, 5, 7, \ldots, 97 \). Each term can be decomposed using partial fractions:

\[
\frac{1}{n \times (n+2)} = \frac{A}{n} + \frac{B}{n+2}
\]

To solve for \( A \) and \( B \), we multiply both sides by \( n \times (n+2) \):

\[
1 = A(n+2) + Bn
\]

Simplifying:

\[
1 = An + 2A + Bn
\]

\[
1 = (A + B)n + 2A
\]

For this equation to hold for all \( n \), the coefficients of \( n \) and the constant term must be equal:

\[
A + B = 0 \quad \text{(coefficient of } n\text{)}
\]
\[
2A = 1 \quad \text{(constant term)}
\]

From \( 2A = 1 \), we get \( A = \frac{1}{2} \). Since \( A + B = 0 \), we have \( B = -\frac{1}{2} \).

Thus,

\[
\frac{1}{n \times (n+2)} = \frac{1/2}{n} - \frac{1/2}{n+2} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right)
\]

Now, substitute this back into the original series:

\[
A = \frac{1}{2} \left( \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{97} - \frac{1}{99}\right) \right)
\]

This is a telescoping series, where most terms cancel out:

\[
A = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{99} \right)
\]

Simplifying this:

\[
A = \frac{1}{2} \times \frac{99 - 3}{3 \times 99} = \frac{96}{2 \times 297} = \frac{48}{297} = \frac{16}{99}
\]

Therefore, the sum of the series \( A \) is:

\[
\boxed{\frac{16}{99}}
\]

Would you like a deeper explanation or have any questions?

Here are 5 related questions to further your understanding:

1. How does the telescoping series help simplify the summation process?
2. Can you generalize this method to similar series with different patterns?
3. What are partial fractions, and how are they useful in series summation?
4. How would the sum change if the series started with a different initial term?
5. Can you find the sum of a similar series, such as \( 1/2 + 1/6 \times 8 + 1/10 \times 12 + \ldots \)?

**Tip:** Recognizing patterns in series and using partial fractions can greatly simplify complex summations.

Explore the summation of a complex series involving partial fractions. Learn how to decompose fractions, apply telescoping series, and calculate sums effectively. Suitable for advanced high school students and beyond.

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