Telescoping Partial‑Fraction Technique

Fast Telescoping Solution

1. Partial‑fraction split

   $$ \frac{3}{(2k-1)(2k+1)} ;=; \frac{A}{2k-1} ;+; \frac{B}{2k+1}. $$

   Solve $3 = A(2k+1)+B(2k-1)$.

   Hence

   $$ \frac{3}{(2k-1)(2k+1)}

   \frac{\tfrac32}{2k-1};-;\frac{\tfrac32}{2k+1}

   \frac{3}{2}!\left(\frac1{2k-1}-\frac1{2k+1}\right). $$

2. Write the sum

   $$ S

   \frac32 \bigl[(1!-!\tfrac13) + (\tfrac13!-!\tfrac15) + (\tfrac15!-!\tfrac17) + \cdots + (\tfrac1{99}!-!\tfrac1{101}) \bigr]. $$

   Everything cancels except the very first and very last terms.

3. Collapse the telescope

   $$ S

   \frac32!\left(1-\frac1{101}\right)

   \frac32\cdot\frac{100}{101}

   \frac{150}{101}. $$