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 \(n\in\mathbb Z^+\).

Video Solution:

Video solution locked

Solution:

Solution locked

Sign in to view the step-by-step solution

Finding similar questions...

Answer:\(\displaystyle\sum_{r=1}^{n}\dfrac{1}{(3r-2)(3r+1)}=\dfrac{n}{3n+1}\)

Need help? Join our JC Math tuition classes.