N2019 P1 Q6

N2019 P1 Q6

Junior College 2
6 marks
Free
  1. By writing \(\dfrac{1}{4r^2-1}\) in partial fractions, find an expression for \(\displaystyle\sum_{r=1}^{n}\dfrac{1}{4r^2-1}\).[4]
  2. Hence find the exact value of \(\displaystyle\sum_{r=1}^{\infty}\dfrac{1}{4r^2-1}\).[2]

Default solution

  1. \(\dfrac{1}{4r^2-1}=\dfrac{1}{(2r-1)(2r+1)}=\dfrac12\left(\dfrac1{2r-1}-\dfrac1{2r+1}\right)\)
    \(S_n=\dfrac12\displaystyle\sum_{r=1}^{n}\left(\dfrac1{2r-1}-\dfrac1{2r+1}\right)\)
    \(S_n=\dfrac12\left(1-\dfrac1{2n+1}\right)=\dfrac{n}{2n+1}\)
  2. \(\displaystyle\sum_{r=1}^{\infty}\dfrac1{4r^2-1}=\lim_{n\to\infty}\dfrac{n}{2n+1}=\dfrac12\)
Answer:(i) \(\dfrac{n}{2n+1}\). (ii) \(\dfrac12\).

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