Multiple choice

The sum of the $n$ terms of the series $\dfrac {1}{3.5}+\dfrac {1}{5.7}+\dfrac {1}{7.9}+....$ is

  1. $\dfrac {n}{3(2n+3)}$
  2. $\dfrac {n+1}{2n+3}$
  3. $\dfrac {n-1}{2n+1}$
  4. $\dfrac {n}{2n+1}$
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A Correct answer
Explanation

The series is sum of 1/((2k+1)(2k+3)). This can be written as (1/2) * (1/(2k+1) - 1/(2k+3)). This is a telescoping series. The sum of n terms is (1/2) * (1/3 - 1/(2n+3)) = (1/2) * ((2n+3-3)/(3(2n+3))) = n / (3(2n+3)).

AI explanation

Use the method of partial fractions to decompose the general term 1 / [(2n+1)(2n+3)] into 1/2 * [1/(2n+1) - 1/(2n+3)]. When summing from 1 to n, the series becomes a telescoping sum where intermediate terms cancel out, leaving 1/2 * [1/3 - 1/(2n+3)]. Simplifying this expression gives 1/2 * [(2n+3 - 3) / 3(2n+3)], which equals n / [3(2n+3)].

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