1/1*3+1/3*5+.......+1/47*49 NEED HELP ASAP!!!!!!!!!!!!!!!!

1/13+1/35+.......+1/4749 NEED HELP ASAP!!!!!!!!!!!!!!!!

If you're working on a mathematics problem involving the sum of the series 1/13 + 1/35 + ... + 1/4749, you're not alone. Many students and math enthusiasts encounter series summation problems that seem complex at first glance but can be simplified through understanding patterns, algebraic manipulation, and known techniques. In this article, we'll explore how to evaluate this particular series step by step, making it easier to understand and solve. Whether you're preparing for exams or just want to improve your series summation skills, this comprehensive guide will help you grasp the concepts thoroughly.

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Understanding the Series: 1/13 + 1/35 + ... + 1/4749

Breaking Down the Series

The series in question is:

\[ S = \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \cdots + \frac{1}{47 \times 49} \]

This sequence involves reciprocals of the product of two odd numbers, starting from 1 and 3, then 3 and 5, and so on, up to 47 and 49.

Identifying the Pattern

Notice that:

• The denominators are products of two consecutive odd numbers.
• The first term is \( \frac{1}{1 \times 3} \).
• The last term is \( \frac{1}{47 \times 49} \).
• The numerator in each term is 1, and the denominators are formed as consecutive odd numbers multiplied together.

The key is to find a way to express each term in a form that simplifies the sum.

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Techniques for Summing the Series

Partial Fraction Decomposition

Partial fraction decomposition is a common technique in summing series involving rational functions. It involves expressing a fraction as a sum of simpler fractions that can cancel out when summed over a sequence.

For the general term:

\[ T_k = \frac{1}{(2k-1)(2k+1)} \]

where \( k \) runs from 1 to 23 (since the last term corresponds to \( 2 \times 23 - 1 = 47 \)).

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Step-by-Step Solution

Expressing the General Term

Let's write the general term explicitly:

\[ T_k = \frac{1}{(2k - 1)(2k + 1)} \]

Our goal is to decompose \( T_k \) into simpler fractions.

Partial Fraction Decomposition

Set:

\[ \frac{1}{(2k - 1)(2k + 1)} = \frac{A}{2k - 1} + \frac{B}{2k + 1} \]

Multiply both sides by \( (2k - 1)(2k + 1) \):

\[ 1 = A(2k + 1) + B(2k - 1) \]

Now, solve for \( A \) and \( B \):

\[ 1 = 2A k + A + 2B k - B \]

Group like terms:

\[ 1 = (2A + 2B)k + (A - B) \]

Since the left side is constant (1), the coefficient of \( k \) on the right must be zero, and the constant term must be 1:

\[
\begin{cases}
2A + 2B = 0 \\
A - B = 1
\end{cases}
\]

Solve the system:

1. From the first equation:

\[ 2A + 2B = 0 \Rightarrow A + B = 0 \Rightarrow B = -A \]

1. Substitute into the second equation:

\[ A - (-A) = 1 \Rightarrow A + A = 1 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2} \]

Then:

\[ B = -A = -\frac{1}{2} \]

So, the decomposition is:

\[ T_k = \frac{1/2}{2k - 1} - \frac{1/2}{2k + 1} \]

or

\[ T_k = \frac{1}{2} \left( \frac{1}{2k - 1} - \frac{1}{2k + 1} \right) \]

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Summing the Series

Expressing the Series as a Telescoping Sum

The original sum:

\[ S = \sum{k=1}^{23} Tk = \frac{1}{2} \sum_{k=1}^{23} \left( \frac{1}{2k - 1} - \frac{1}{2k + 1} \right) \]

Break into two sums:

\[ S = \frac{1}{2} \left( \sum{k=1}^{23} \frac{1}{2k - 1} - \sum{k=1}^{23} \frac{1}{2k + 1} \right) \]

Let's write out the terms explicitly:

• First sum:

\[ \sum_{k=1}^{23} \frac{1}{2k - 1} = \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{45} + \frac{1}{47} \]

• Second sum:

\[ \sum_{k=1}^{23} \frac{1}{2k + 1} = \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots + \frac{1}{47} + \frac{1}{49} \]

Notice that many terms cancel when we subtract the sums:

\[ S = \frac{1}{2} \left( \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{45} + \frac{1}{47}\right) - \left(\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots + \frac{1}{47} + \frac{1}{49}\right) \right) \]

After cancellation:

\[ S = \frac{1}{2} \left( 1 - \frac{1}{49} \right) \]

because all intermediate terms \( \frac{1}{3}, \frac{1}{5}, \dots, \frac{1}{47} \) cancel out.

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Final Calculation and Result

Calculate:

\[ 1 - \frac{1}{49} = \frac{49}{49} - \frac{1}{49} = \frac{48}{49} \]

Therefore,

\[ S = \frac{1}{2} \times \frac{48}{49} = \frac{48}{98} \]

Simplify numerator and denominator:

\[ \frac{48}{98} = \frac{24}{49} \]

Final answer:

\[
\boxed{
\frac{24}{49}
}
\]

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Summary and Key Takeaways

• Expressing the general term using partial fractions simplifies the summation process.
• Recognizing telescoping sums can drastically reduce the complexity of series calculations.
• Canceling common terms in telescoping series leaves only the first and last terms, making the sum straightforward to evaluate.
• The sum of the series 1/13 + 1/35 + ... + 1/4749 is \(\frac{24}{49}\).

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Additional Tips for Series Summation

Practice Recognizing Patterns

Many series involve patterns that can be exploited through algebraic manipulation. Always look for opportunities to rewrite terms in a form conducive to telescoping or partial fractions.

Verify with Smaller Examples

Before tackling large series, test your approach on smaller series to ensure your method works.

Use Known Series and Identities

Familiarity with harmonic series, telescoping series, and partial fraction identities can accelerate problem-solving.

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If you're struggling with series summations or need further help, consider consulting math textbooks, online tutorials, or seeking assistance from teachers or tutors. Mastery of these techniques will significantly improve your problem-solving skills and confidence in tackling complex series.

Remember: Practice makes perfect, and breaking down complex series into manageable parts is the key to solving them efficiently.