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Question Number 174377 by AgniMath last updated on 31/Jul/22

Answered by okbruh123 last updated on 31/Jul/22

$$\mathrm{ez}\mathrm{telescoping}\mathrm{lol}$$$$2+\frac{1}{3\cdot 5}+...+\frac{1}{11\cdot 13}$$$$2+\frac{1}{2}(\frac{5-3}{3\cdot 5}+...+\frac{13-11}{11\cdot 13})$$$$2+\frac{1}{2}(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13})$$$$2+\frac{1}{2}(\frac{1}{3}-\frac{1}{13})=2+\frac{1}{2}\left(\frac{10}{39}\right)=2+\frac{5}{39}\mathrm{now}\mathrm{do}\mathrm{basic}\mathrm{maths}$$

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