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Question Number 7903 by tawakalitu last updated on 24/Sep/16

Commented by sou1618 last updated on 24/Sep/16

$$S=\frac{\underset{k=1}{\sum ^{99}}\sqrt{10+\sqrt{k}}}{\underset{k=1}{\sum ^{99}}\sqrt{10-\sqrt{k}}}$$$$S\fallingdotseq S{}^{\prime }=\frac{{\int }_0^{100}\sqrt{10+\sqrt{x}}dx}{{\int }_0^{100}\sqrt{10-\sqrt{y}}dy}$$$$t=10+\sqrt{x}$$$$x={(t-10)}^2$$$$dx=2t-20dt$$$$x(0\to 100)\iff t(10\to 20)$$$$$$$$u=10-\sqrt{y}$$$$y={(10-u)}^2$$$$dy=2t-20du$$$$y(0\to 100)\iff u(10\to 0)$$$$$$$$S{}^{\prime }=\frac{{\int }_{10}^{20}\sqrt{10+\sqrt{{(t-10)}^2}}(2t-20)dt}{{\int }_{10}^0\sqrt{10-\sqrt{{(10-u)}^2}}(2u+20)du}$$$$=\frac{{\int }_{10}^{20}\sqrt{10+\mid t-10\mid }(2t-20)dt}{{\int }_{10}^0\sqrt{10-\mid 10-u\mid }(2u-20)du}$$$$=\frac{{\int }_{10}^{20}2\sqrt{t}(t-10)dt}{{\int }_{10}^{0}2\sqrt{u}(u-10)du}$$$$=\frac{{\int }_{10}^{20}\sqrt{t^3}-10\sqrt{t}dt}{{\int }_{10}^0\sqrt{u^3}-10\sqrt{u}du}$$$$=\frac{{[\frac{2}{5}\sqrt{t^5}-10\frac{2}{3}\sqrt{t^3}]}_{10}^{20}}{{[\frac{2}{5}\sqrt{u^5}-10\frac{2}{3}\sqrt{u^3}]}_{10}^0}$$$$=\frac{\frac{2}{5}(\sqrt{{20}^5}-\sqrt{{10}^5})-10\frac{2}{3}(\sqrt{{20}^3}-\sqrt{{10}^3})}{0-(\frac{2}{5}\sqrt{{10}^5}-10\frac{2}{3}\sqrt{{10}^3})}\times \frac{3\sqrt{10}}{3\sqrt{10}}$$$$=\frac{\frac{6}{5}(4000\sqrt{2}-1000)-20(200\sqrt{2}-100)}{-\frac{6}{5}\left(1000\right)+20\left(100\right)}$$$$=\frac{4800\sqrt{2}-1200-4000\sqrt{2}+2000}{-1200+2000}$$$$=\frac{800\sqrt{2}+800}{800}$$$$=1+\sqrt{2}$$$$so$$$$S\fallingdotseq 1+\sqrt{2}$$

Commented by tawakalitu last updated on 24/Sep/16

$$Wowthankssomuch.ireallyappreciate.$$

Commented by Yozzia last updated on 24/Sep/16

$$HowcomeS\fallingdotseq S{}^{\prime }?$$

Commented by sou1618 last updated on 24/Sep/16

$$\sqrt{10\pm \sqrt{k}}\fallingdotseq {\int }_{k-1}^k\sqrt{10\pm \sqrt{x}}dx$$$$\underset{k=1}{\sum ^{100}}\sqrt{10\pm \sqrt{k}}\fallingdotseq \underset{k=1}{\sum ^{100}}{\int }_{k-1}^k\sqrt{10\pm \sqrt{x}}dx$$$$\left(\underset{k=1}{\sum ^{99}}\sqrt{10\pm \sqrt{k}}\right)+\sqrt{10\pm 10}\fallingdotseq {\int }_0^{100}\sqrt{10\pm \sqrt{x}}dx$$$$now$$$$\underset{k=1}{\sum ^{99}}\sqrt{10\pm \sqrt{k}}>>\sqrt{10\pm 10}$$$$Ithink\sqrt{10+10}islessthan\sqrt{2}\mathrm{\%}of\mathrm{\Sigma }$$$$because...when0<k<100$$$$\sqrt{2}\sqrt{10+\sqrt{k}}>\sqrt{10+10}$$$$$$$$$$$$\underset{k=1}{\sum ^{99}}\sqrt{10\pm \sqrt{k}}\fallingdotseq {\int }_0^{100}\sqrt{10\pm \sqrt{x}}dx$$$$so$$$$S\fallingdotseq S{}^{\prime }$$

Answered by prakash jain last updated on 02/Oct/16

$$\mathrm{answer}\mathrm{n}\mathrm{comments}$$

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