N2022 P2 Q4

N2022 P2 Q4

Junior College 2
9 marks
Free
  1. Write \(\frac{1}{9r^2+3r-2}\) in partial fractions.[2]
  2. Find an expression in terms of \(m\) for \(\sum_{r=m}^{3m}\left(\frac{1}{9r^2+3r-2}\right)\). Write your answer as a single fraction in terms of \(m\).[4]
  3. Find \(\sum_{r=1}^{\infty}\left(\frac{1}{9r^2+3r-2}\right)\).[1]
  4. Find the smallest value of \(n\) for which \(\sum_{r=1}^{n}\left(\frac{1}{9r^2+3r-2}\right)\) differs from \(\sum_{r=1}^{\infty}\left(\frac{1}{9r^2+3r-2}\right)\) by less than 0.004.[2]

Default solution

  1. \(9r^2+3r-2=(3r-1)(3r+2)\)
    \(\frac{1}{9r^2+3r-2}=\frac{1}{3(3r-1)}-\frac{1}{3(3r+2)}\)
  2. \(\sum_{r=m}^{3m}\frac{1}{9r^2+3r-2}=\frac13\left(\frac1{3m-1}-\frac1{9m+2}\right)\) (telescoping)
    \(=\frac{2m+1}{(3m-1)(9m+2)}\)
  3. \(\sum_{r=1}^{N}\frac{1}{9r^2+3r-2}=\frac13\left(\frac12-\frac1{3N+2}\right)\)
    \(N\to\infty\Rightarrow S_\infty=\frac16\)
  4. \(S_\infty-S_n=\frac{1}{3(3n+2)}=\frac1{9n+6}<0.004=\frac1{250}\)
    \(9n+6>250\Rightarrow n>\frac{244}{9}\)
    \(n=27:\ \frac1{249}>0.004;\quad n=28:\ \frac1{258}<0.004\)
    \(\therefore\ n=28\)
Answer:(a) \(\frac{1}{3(3r-1)}-\frac{1}{3(3r+2)}\) (b) \(\frac{2m+1}{(3m-1)(9m+2)}\) (c) \(\frac16\) (d) \(28\)

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