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Question Number 149189 by Naser last updated on 03/Aug/21

Commented by Tawa11 last updated on 03/Aug/21

$$\mathrm{great}$$

Answered by Ar Brandon last updated on 03/Aug/21

$$S=\underset{k=1}{\sum ^n}\frac{1}{(3k-2)(2k+1)}$$$$=\underset{k=1}{\sum ^n}(\frac{3}{7}\cdot \frac{1}{(3k-2)}-\frac{2}{7}\cdot \frac{1}{(2k+1)})$$$$=\frac{1}{7}\underset{k=1}{\sum ^n}\frac{1}{k-\frac{2}{3}}-\frac{1}{k+\frac{1}{2}}$$

Answered by nimnim last updated on 03/Aug/21

$$Letmegiveatry.$$$$S=\underset{k=1}{\sum ^n}\frac{1}{(3k-2)(2k+1)}$$$$=\underset{k=1}{\sum ^n}(\frac{3}{7}.\frac{1}{3k-2}-\frac{2}{7}.\frac{1}{2k+1})$$$$=\frac{3}{7}\underset{k=1}{\sum ^n}\frac{1}{(3k-2)}-\frac{2}{7}\underset{k=1}{\sum ^n}\frac{1}{(2k+1)}$$$$=\frac{3}{7}(\frac{1}{1}+\frac{1}{4}+\frac{1}{7}+....+\frac{1}{1+(n-1).3})$$$$-\frac{2}{7}(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{3+(n-1).2})$$$$bothareharmonicprogressiontonterms$$$$S=\frac{1}{d}ln\mid \left(\frac{2a+(2n-1)d}{2a-d}\right)\mid$$$$\therefore S=\frac{3}{7}\times \frac{1}{3}ln\mid \left(\frac{2.1+(2n-1).3}{2.1-3}\right)\mid -\frac{2}{7}\times \frac{1}{2}ln\mid \left(\frac{2.3+(2n-1).2}{2.3-2}\right)\mid$$$$=\frac{1}{7}ln\mid \frac{2n-1}{-1}\mid -\frac{1}{7}ln\mid \frac{4+4n}{4}\mid$$$$=\frac{1}{7}(ln\mid 1-2n\mid -ln\mid 1+n\mid )$$$$=\frac{1}{7}ln\mid \frac{1-2n}{1+n}\mid$$$$$$

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