Question

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...$

Answer 1

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...Wehave,Let{T}_r\text{be the rth term of the given series. Then,}{T}_r=\frac{1}{(3r-2)(3r+1)},r=1,2,....,n\Rightarrow T_r=\frac{1}{3}[\frac{1}{3r-2}-\frac{1}{3r+1}]\therefore requiredsum=\frac{1}{3}{\sum }_{r-1}^n{T}_r=\frac{1}{3}{\sum }_{r-1}^n[\frac{1}{3n-2}-\frac{1}{3n+1}]=\frac{1}{3}[(1-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})...\frac{1}{3n-2}-\frac{1}{3n+1}]=\frac{1}{2}[1-\frac{1}{3n+1}]=\frac{1}{3}\left[\frac{3n+1-1}{3n+1}\right]=\frac{1}{3}\left[\frac{3n+1-1}{3n+1}\right]=\frac{1}{3}\times \frac{3n}{3n+1}=\frac{n}{3n+1}Hence,requiredsum=\frac{n}{3n+1}$