\(\sum\limits_{x=1}^{99}\frac{1}{x(x+1)}\) = ____________.
2015
\(\sum\limits_{x=1}^{99}\frac{1}{x(x+1)}\) = ____________.
Attempted by 59 students.
Show answer & explanation
Correct answer: 0.99
Key idea: rewrite the term as a difference so the sum telescopes.
Rewrite the general term: 1/(x(x+1)) = 1/x - 1/(x+1).
Apply this inside the sum: sum from x=1 to 99 of (1/x - 1/(x+1)).
Write the first and last few terms to see cancellation: (1 - 1/2) + (1/2 - 1/3) + ... + (1/99 - 1/100).
All intermediate terms cancel, leaving 1 - 1/100 = 99/100, which equals 0.99.