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Question Number 188151 by Shrinava last updated on 26/Feb/23

$$\mathrm{Simplify}:$$$$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{{2022}^2}+\frac{1}{{2023}^2}}$$

Answered by BaliramKumar last updated on 27/Feb/23

$$Solution:-$$$$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{{2022}^2}+\frac{1}{{2023}^2}}$$$$t_1=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=\sqrt{1+1+\frac{1}{4}}=\sqrt{\frac{9}{4}}=\frac{3}{2}$$$$t_2=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=\sqrt{1+\frac{1}{4}+\frac{1}{9}}=\sqrt{\frac{49}{36}}=\frac{7}{6}$$$$t_3=\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}=\sqrt{1+\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{169}{144}}=\frac{13}{12}$$$$t_4..............$$$$t_{5..............}$$$$t_{2022}=\sqrt{1+\frac{1}{{2022}^2}+\frac{1}{{2023}^2}}$$$$S_1=t_1=\frac{3}{2}=1+\frac{1}{2}=1\frac{1}{2}$$$$S_2=S_1+t_2=\frac{3}{2}+\frac{7}{6}=\frac{16}{6}=\frac{8}{3}=2+\frac{2}{3}=2\frac{2}{3}$$$$S_3=S_2+t_3=\frac{8}{3}+\frac{13}{12}=\frac{45}{12}=\frac{15}{4}=3+\frac{3}{4}=3\frac{3}{4}$$$$S_{4..........}$$$$S_{2022}=2022+\frac{2022}{2023}=\overline{)\begin{array}{c}2022\frac{2022}{2023}\end{array}}\mathrm{Answer}$$

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