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

Answer: (B)

5

Answer 2

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).