Question Number 213413 by hardmath last updated on 04/Nov/24
![Find: A= [(√1)] + [(√2)] + [(√3) ]+...+ [(√(323))] = ?](https://www.tinkutara.com/question/Q213413.png)
$$\mathrm{Find}: \\ $$$$\mathrm{A}=\:\left[\sqrt{\mathrm{1}}\right]\:+\:\left[\sqrt{\mathrm{2}}\right]\:+\:\left[\sqrt{\mathrm{3}}\:\right]+…+\:\left[\sqrt{\mathrm{323}}\right]\:=\:? \\ $$
Answered by mehdee7396 last updated on 04/Nov/24
![=([(√1)]+[(√2)]+[(√3)]+)+([4]+[5]+...+[[(√8)]) +([(√9)]+[(√(10))]+...+[(√(24))])+... +([(√(289))]+[(√(290))]+...+[(√(323))]) =3(1)+5(2)+7(3)+...+35(17) =Σ_(n=1) ^(17) (2n+1)n=2Σ_(n=1) ^(17) n^2 +Σ_(n=1) ^(17) n =2×((17×18×35)/6)+((17×18)/2) =3723](https://www.tinkutara.com/question/Q213415.png)
$$=\left(\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+\right)+\left(\left[\mathrm{4}\right]+\left[\mathrm{5}\right]+…+\left[\left[\sqrt{\mathrm{8}}\right]\right)\right. \\ $$$$+\left(\left[\sqrt{\mathrm{9}}\right]+\left[\sqrt{\mathrm{10}}\right]+…+\left[\sqrt{\mathrm{24}}\right]\right)+… \\ $$$$+\left(\left[\sqrt{\mathrm{289}}\right]+\left[\sqrt{\mathrm{290}}\right]+…+\left[\sqrt{\mathrm{323}}\right]\right) \\ $$$$=\mathrm{3}\left(\mathrm{1}\right)+\mathrm{5}\left(\mathrm{2}\right)+\mathrm{7}\left(\mathrm{3}\right)+…+\mathrm{35}\left(\mathrm{17}\right) \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{17}} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right){n}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\mathrm{17}} {\sum}}{n}^{\mathrm{2}} +\underset{{n}=\mathrm{1}} {\overset{\mathrm{17}} {\sum}}{n} \\ $$$$=\mathrm{2}×\frac{\mathrm{17}×\mathrm{18}×\mathrm{35}}{\mathrm{6}}+\frac{\mathrm{17}×\mathrm{18}}{\mathrm{2}} \\ $$$$=\mathrm{3723}\: \\ $$$$ \\ $$
Commented by hardmath last updated on 04/Nov/24
$$ \\ $$dear friend, Is there an easier way to do this? I didn't quite understand it
Commented by hardmath last updated on 05/Nov/24
$$\mathrm{2}\:\centerdot\:\frac{\mathrm{17}\centerdot\mathrm{18}\centerdot\mathrm{35}?}{\mathrm{6}}\:\:\:\mathrm{dear}\:\mathrm{ser}\:\mathrm{how}? \\ $$
Answered by Frix last updated on 05/Nov/24
$$\lfloor\sqrt{{n}}\rfloor=\mathrm{1},\:\mathrm{1}\leqslant{n}\leqslant\mathrm{3}\:\Rightarrow\:\mathrm{3}×\mathrm{1} \\ $$$$\lfloor\sqrt{{n}}\rfloor=\mathrm{2},\:\mathrm{4}\leqslant{n}\leqslant\mathrm{8}\:\Rightarrow\:\mathrm{5}×\mathrm{2} \\ $$$$\lfloor\sqrt{{n}}\rfloor=\mathrm{3},\:\mathrm{9}\leqslant{n}\leqslant\mathrm{15}\:\Rightarrow\:\mathrm{7}×\mathrm{3} \\ $$$$… \\ $$$$\lfloor\sqrt{{n}}\rfloor={k},\:{k}^{\mathrm{2}} \leqslant{n}\leqslant\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}={k}^{\mathrm{2}} +\mathrm{2}{k}\:\Rightarrow\:\left(\mathrm{2}{k}+\mathrm{1}\right)×{k} \\ $$$$\mathrm{323}=\mathrm{18}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\:\mathrm{1}\leqslant{k}\leqslant\mathrm{17} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right){k}=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{k}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{k}= \\ $$$$=\frac{{m}\left({m}+\mathrm{1}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)}{\mathrm{3}}+\frac{{m}\left({m}+\mathrm{1}\right)}{\mathrm{2}}= \\ $$$$=\frac{{m}\left({m}+\mathrm{1}\right)\left(\mathrm{4}{m}+\mathrm{5}\right)}{\mathrm{6}} \\ $$$${m}=\mathrm{17}\:\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{17}×\mathrm{18}×\mathrm{73}}{\mathrm{6}}=\mathrm{3723} \\ $$
Commented by hardmath last updated on 05/Nov/24
$$\mathrm{thankyoudear}\:\mathrm{professor} \\ $$