Question

1/(3² - 1) + 1/(5² - 1) + 1/(7² - 1) + ⋯ + 1/(201)² - 1 is equal to :

• A. 25/101
• B. 101/408
• C. 99/400
• D. 101/404

Hint:

Telescoping sums are the most efficient way to sum series involving reciprocal quadratics.

Answer 1

The Correct Option is A

Step 1: General term $T_r = \frac{1}{(2r+1)^2 - 1} = \frac{1}{(2r+1-1)(2r+1+1)} = \frac{1}{2r(2r+2)} = \frac{1}{4r(r+1)}$.
Step 2: Use partial fractions: $T_r = \frac{1}{4} \left[ \frac{1}{r} - \frac{1}{r+1} \right]$.
Step 3: The series starts from $3^2-1$ ($r=1$) to $201^2-1$ ($r=100$).
Step 4: Sum $= \frac{1}{4} \sum_{r=1}^{100} \left( \frac{1}{r} - \frac{1}{r+1} \right) = \frac{1}{4} \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ....... + (\frac{1}{100} - \frac{1}{101}) \right]$.
Step 5: Sum $= \frac{1}{4} [1 - \frac{1}{101}] = \frac{1}{4} \times \frac{100}{101} = \frac{25}{101}$.