Question

Evaluate the sum: \[ \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + \dots \text{ up to 50 terms=} \]

• A. \( \frac{50}{203} \)
• B. \( \frac{50}{609} \)
• C. \( \frac{150}{203} \)
• D. \( \frac{25}{609} \)

Hint:

For summations involving product terms in the denominator, consider partial fraction decomposition to simplify the series into a telescoping sum, making evaluation straightforward.

Answer 1

The Correct Option is B

The given series is: \[ S = \sum_{n=1}^{50} \frac{1}{(4n - 1)(4n + 3)}. \] Step 1: Partial Fraction Decomposition
We express each term using partial fractions: \[ \frac{1}{(4n - 1)(4n + 3)} = \frac{A}{4n - 1} + \frac{B}{4n + 3}. \] Multiplying both sides by \( (4n - 1)(4n + 3) \), we get: \[ 1 = A(4n + 3) + B(4n - 1). \] Expanding and equating coefficients: \[ 1 = (4A + 4B)n + (3A - B). \] Solving for \( A \) and \( B \): \[ 4A + 4B = 0, \quad 3A - B = 1. \] From the first equation, \( A + B = 0 \Rightarrow B = -A \). Substituting in the second equation: \[ 3A - (-A) = 1 \Rightarrow 4A = 1 \Rightarrow A = \frac{1}{4}, \quad B = -\frac{1}{4}. \] Step 2: Telescoping the Series
Thus, rewriting the general term: \[ \frac{1}{(4n - 1)(4n + 3)} = \frac{1}{4} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right). \] Summing over 50 terms: \[ S = \frac{1}{4} \sum_{n=1}^{50} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right). \] This forms a telescoping series, where most terms cancel, leaving: \[ S = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{203} \right). \] Step 3: Evaluating the Final Expression
\[ S = \frac{1}{4} \times \frac{203 - 3}{3 \times 203} = \frac{1}{4} \times \frac{200}{609} = \frac{50}{609}. \] Thus, the sum of the series is: \[ \boxed{\frac{50}{609}} \]

Answer 2

Evaluate the sum:
\[ S = \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + \dots \text{ up to 50 terms} \]

Step 1: Observe the pattern of denominators:
Each term is of the form:
\[ T_n = \frac{1}{(4n - 1)(4n + 3)}, \quad n = 1, 2, \dots, 50 \]

Step 2: Use partial fraction decomposition:
\[ \frac{1}{(4n - 1)(4n + 3)} = \frac{A}{4n - 1} + \frac{B}{4n + 3} \]
Multiply both sides by the denominator:
\[ 1 = A(4n + 3) + B(4n - 1) \]

Step 3: Equate coefficients:
\[ 1 = (4A + 4B)n + (3A - B) \] For this to hold for all \( n \), coefficients must satisfy:
\[ 4A + 4B = 0 \implies A = -B \] \[ 3A - B = 1 \] Substitute \( A = -B \):
\[ 3(-B) - B = 1 \implies -3B - B = 1 \implies -4B = 1 \implies B = -\frac{1}{4} \] \[ A = -B = \frac{1}{4} \]

Step 4: So,
\[ T_n = \frac{1/4}{4n - 1} - \frac{1/4}{4n + 3} = \frac{1}{4} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right) \]

Step 5: Write the sum:
\[ S = \frac{1}{4} \sum_{n=1}^{50} \left( \frac{1}{4n - 1} - \frac{1}{4n + 3} \right) \]

Step 6: Write out the first few terms:
\[ = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \frac{1}{11} - \frac{1}{15} + \dots + \frac{1}{199} - \frac{1}{203} \right) \]

Step 7: Notice telescoping cancellation:
\[ S = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{203} \right) = \frac{1}{4} \times \frac{203 - 3}{3 \times 203} = \frac{1}{4} \times \frac{200}{609} = \frac{200}{2436} = \frac{50}{609} \]

Therefore, the sum is:
\[ \boxed{\frac{50}{609}} \]