Given that $\displaystyle \sum_{r=1}^{n}\frac{1}{(3r-1)(3r+2)}=\frac{1}{6}-\frac{1}{3(3n+2)}$

Question

Given that $\displaystyle \sum_{r=1}^{n}\frac{1}{(3r-1)(3r+2)}=\frac{1}{6}-\frac{1}{3(3n+2)}$, find $\displaystyle \sum_{r=4}^{n}\frac{1}{(3r+2)(3r+5)}$ in terms of $n$.[4]

Accepted Answer

$\frac{n-3}{14(3n+5)}$

Shifted rational sum with omitted terms

Shifted rational sum with omitted terms

Video Solution:

Solution:

Solution locked