Question

If the sum of the series $$ \frac{1}{1 \cdot (1 + d)} + \frac{1}{(1 + d)(1 + 2d)} + \cdots + \frac{1}{(1 + 9d)(1 + 10d)} $$ is equal to 5, then \(50d\) is equal to: 

• A. 20
• B. 5
• C. 15
• D. 10

Answer 1

The Correct Option is B

Step 1: General term of the series
The general term of the given series is: \[ T_n = \frac{1}{(1 + (n-1)d)(1 + nd)}. \]

Using partial fraction decomposition: \[ \frac{1}{(1 + (n-1)d)(1 + nd)} = \frac{A}{1 + (n-1)d} + \frac{B}{1 + nd}. \] 

Simplify: \[ \frac{1}{(1 + (n-1)d)(1 + nd)} = \frac{A(1 + nd) + B(1 + (n-1)d)}{(1 + (n-1)d)(1 + nd)}. \] 

Expanding numerators: \[ 1 = A(1 + nd) + B(1 + (n-1)d). \] 

Equating numerators: \[ 1 = A + And + B + Bnd - Bd. \] 

Combine terms: \[ 1 = (A + B) + (Ad + Bd)n - Bd. \] 

Equating coefficients: 
1. \(A + B = 0\), 
2. \(Ad + Bd = 0\), 
3. \(-Bd = 1\).
From \(A + B = 0\): \[ B = -A. \] Substitute \(B = -A\) into \(-Bd = 1\): \[ -(-A)d = 1 \quad \implies \quad A = \frac{1}{d}. \] Thus, \(B = -\frac{1}{d}\).

The partial fraction decomposition becomes: \[ \frac{1}{(1 + (n-1)d)(1 + nd)} = \frac{1}{d} \left[ \frac{1}{1 + (n-1)d} - \frac{1}{1 + nd} \right]. \]

Step 2: Simplify the series
The series becomes: \[ \sum_{n=1}^{10} \frac{1}{(1 + (n-1)d)(1 + nd)} = \frac{1}{d} \left[ 1 - \frac{1}{1 + 10d} \right]. \] 

Simplify: \[ \text{Sum} = \frac{1}{d} \times \frac{(1 + 10d) - 1}{1 + 10d}. \] 

Combine terms: \[ \text{Sum} = \frac{1}{d} \times \frac{10d}{1 + 10d}. \]

Step 3: Solve for \(d\)
Given that the sum of the series is 5: \[ \frac{10}{1 + 10d} = 5. \] 

Simplify: \[ 1 + 10d = 2 \quad \implies \quad 10d = 1 \quad \implies \quad d = \frac{1}{10}. \]

Step 4: Compute \(50d\)
\[ 50d = 50 \times \frac{1}{10} = 5. \]

Final Answer is Option (2).