Question

If $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..............+\frac{1}{(3n-1)(3n+2)}=\frac{n}{an+4}.$ Find $a$

Answer 1

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1)(3n+2)}=\frac{n}{an+4}\frac{1}{3}(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{(3n-1)(3n+2)})=\frac{n}{an+4}\frac{1}{3}((\frac{1}{2}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11})+....+(\frac{1}{3n-1}-\frac{1}{3n+2}))=\frac{n}{an+4}\frac{1}{3}(\frac{1}{2}-\frac{1}{3n+2})=\frac{n}{an+4}\frac{1}{3}\left(\frac{3n}{6n+4}\right)=\frac{n}{an+4}\frac{n}{6n+4}=\frac{n}{an+4}a=6$