A Telescoping Reciprocal Sum
A Telescoping Reciprocal Sum
IB Year 5 | Grade 11
Prove by mathematical induction that \(\displaystyle\sum_{r=1}^{n}\dfrac{1}{(3r-2)(3r+1)}=\dfrac{n}{3n+1}\) for every \(n\in\mathbb Z^+\).
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Answer:\(\displaystyle\sum_{r=1}^{n}\dfrac{1}{(3r-2)(3r+1)}=\dfrac{n}{3n+1}\)
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