Question Number 157891 by HongKing last updated on 29/Oct/21
Answered by TheSupreme last updated on 29/Oct/21
$$\left(\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+\frac{\mathrm{7}}{\mathrm{144}}+\frac{\mathrm{9}}{\mathrm{400}}+\frac{\mathrm{11}}{\mathrm{900}}\right){x}<\mathrm{70} \\ $$$$\left(\frac{\mathrm{2400}+\mathrm{500}+\mathrm{175}+\mathrm{81}+\mathrm{44}}{\mathrm{3600}}\right){x}<\mathrm{70} \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{9}}\right){x}<\mathrm{70} \\ $$$${x}<\frac{\mathrm{630}}{\mathrm{8}}=\mathrm{78}.\mathrm{75} \\ $$$${largest}\:{x}\:=\:\mathrm{78} \\ $$$$ \\ $$$$ \\ $$
Commented by HongKing last updated on 29/Oct/21
$$\left.\mathrm{a}\left.\right)\left.\mathrm{7}\left.\mathrm{4}\left.\:\:\:\mathrm{b}\right)\mathrm{71}\:\:\:\mathrm{c}\right)\mathrm{69}\:\:\:\mathrm{d}\right)\mathrm{64}\:\:\:\mathrm{e}\right)\mathrm{55} \\ $$
Answered by mr W last updated on 29/Oct/21
$$\frac{\mathrm{2}{k}+\mathrm{1}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Sigma=\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)+\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)+…+\left(\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{36}}=\frac{\mathrm{35}}{\mathrm{36}} \\ $$$$\frac{\mathrm{35}}{\mathrm{36}}{x}<\mathrm{70} \\ $$$${x}<\frac{\mathrm{70}×\mathrm{36}}{\mathrm{35}}=\mathrm{72} \\ $$$$\Rightarrow{x}_{{max}} =\mathrm{71}\:\Rightarrow{answer}\:{b} \\ $$
Commented by HongKing last updated on 29/Oct/21
$$\mathrm{alot}\:\mathrm{thankyou}\:\mathrm{ser}\:\mathrm{cool} \\ $$