Question Number 190067 by DAVONG last updated on 26/Mar/23
Answered by mehdee42 last updated on 26/Mar/23
![S=[(√1)]+[(√2)]+...+[(√(500))] [(√1)]+[(√2)]+[(√3)]=3×1 [(√4)]+[(√5)]+[6]+[(√7)]+[(√8)]=5×2 ⋮ [(√(441))]+[(√(442))]+...+[(√(483))]=43×21 [(√(484))]+[(√(485))]+...+[(√(500))]=27×22 ⇒S=Σ_1 ^(21) (2i+1)i+27×22=2Σ_1 ^(21) i^2 +Σ_1 ^(21) i+374 =2×((21×22×43)/6)+((21×22)/2)+374=7227](https://www.tinkutara.com/question/Q190070.png)
$${S}=\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+…+\left[\sqrt{\mathrm{500}}\right] \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]=\mathrm{3}×\mathrm{1} \\ $$$$\left[\sqrt{\mathrm{4}}\right]+\left[\sqrt{\mathrm{5}}\right]+\left[\mathrm{6}\right]+\left[\sqrt{\mathrm{7}}\right]+\left[\sqrt{\mathrm{8}}\right]=\mathrm{5}×\mathrm{2} \\ $$$$\vdots \\ $$$$\left[\sqrt{\mathrm{441}}\right]+\left[\sqrt{\mathrm{442}}\right]+…+\left[\sqrt{\mathrm{483}}\right]=\mathrm{43}×\mathrm{21} \\ $$$$\left[\sqrt{\mathrm{484}}\right]+\left[\sqrt{\mathrm{485}}\right]+…+\left[\sqrt{\mathrm{500}}\right]=\mathrm{27}×\mathrm{22} \\ $$$$\Rightarrow{S}=\underset{\mathrm{1}} {\overset{\mathrm{21}} {\sum}}\left(\mathrm{2}{i}+\mathrm{1}\right){i}+\mathrm{27}×\mathrm{22}=\mathrm{2}\underset{\mathrm{1}} {\overset{\mathrm{21}} {\sum}}{i}^{\mathrm{2}} +\underset{\mathrm{1}} {\overset{\mathrm{21}} {\sum}}{i}+\mathrm{374} \\ $$$$=\mathrm{2}×\frac{\mathrm{21}×\mathrm{22}×\mathrm{43}}{\mathrm{6}}+\frac{\mathrm{21}×\mathrm{22}}{\mathrm{2}}+\mathrm{374}=\mathrm{7227} \\ $$
Commented by DAVONG last updated on 27/Mar/23
$$\mathrm{Thanks}\:\mathrm{sire}\:! \\ $$
Answered by mr W last updated on 26/Mar/23
![S_n =[(√1)]+[(√2)]+[(√3)]+...+[(√n)] let m=⌊(√n)⌋ [(√1)]+[(√2)]+[(√3)]=3×1 [(√4)]+[(√5)]+[(√6)]+[(√7)]+[(√8)]=5×2 [(√9)]+[(√(10))]+...+[(√(15))]=7×3 ... [(√k^2 )]+[(√(k^2 +1))]+...+[(√(k^2 +2k))]=(2k+1)×k ... [(√m^2 )]+[(√(m^2 +1))]+...+(√n)=(n+1−m^2 )×m S_n =Σ_(k=1) ^(m−1) (2k+1)k+(n+1−m^2 )m S_n =2Σ_(k=1) ^(m−1) k^2 +Σ_(k=1) ^(m−1) k+(n+1−m^2 )m S_n =2×(((m−1)m(2m−1))/6)+(((m−1)m)/2)+(n+1−m^2 )m S_n =(((m−1)m(4m+1))/6)+(n+1−m^2 )m S_n =m(n+1)−((m(m+1)(2m+1))/6) with n=500: m=⌊(√(500))⌋=22 S_(500) =22×501−((22×23×45)/6)=7227 ===================== [(√1)]+[(√2)]+[(√3)]+...+[(√n)]=m(n+1)−((m(m+1)(2m+1))/6) i.e. [(√1)]+[(√2)]+[(√3)]+...+[(√n)]=m(n+1)−Σ_(k=1) ^m k^2 generally: [(1)^(1/r) ]+[(2)^(1/r) ]+[(3)^(1/r) ]+...+[(n)^(1/r) ]=m(n+1)−Σ_(k=1) ^m k^r with m=⌊(n)^(1/r) ⌋](https://www.tinkutara.com/question/Q190072.png)
$${S}_{{n}} =\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{{n}}\right] \\ $$$${let}\:{m}=\lfloor\sqrt{{n}}\rfloor \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]=\mathrm{3}×\mathrm{1} \\ $$$$\left[\sqrt{\mathrm{4}}\right]+\left[\sqrt{\mathrm{5}}\right]+\left[\sqrt{\mathrm{6}}\right]+\left[\sqrt{\mathrm{7}}\right]+\left[\sqrt{\mathrm{8}}\right]=\mathrm{5}×\mathrm{2} \\ $$$$\left[\sqrt{\mathrm{9}}\right]+\left[\sqrt{\mathrm{10}}\right]+…+\left[\sqrt{\mathrm{15}}\right]=\mathrm{7}×\mathrm{3} \\ $$$$… \\ $$$$\left[\sqrt{{k}^{\mathrm{2}} }\right]+\left[\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\right]+…+\left[\sqrt{{k}^{\mathrm{2}} +\mathrm{2}{k}}\right]=\left(\mathrm{2}{k}+\mathrm{1}\right)×{k} \\ $$$$… \\ $$$$\left[\sqrt{{m}^{\mathrm{2}} }\right]+\left[\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right]+…+\sqrt{{n}}=\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right)×{m} \\ $$$$ \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right){k}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} =\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\sum}}{k}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\sum}}{k}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} =\mathrm{2}×\frac{\left({m}−\mathrm{1}\right){m}\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{6}}+\frac{\left({m}−\mathrm{1}\right){m}}{\mathrm{2}}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} =\frac{\left({m}−\mathrm{1}\right){m}\left(\mathrm{4}{m}+\mathrm{1}\right)}{\mathrm{6}}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} ={m}\left({n}+\mathrm{1}\right)−\frac{{m}\left({m}+\mathrm{1}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${with}\:{n}=\mathrm{500}: \\ $$$${m}=\lfloor\sqrt{\mathrm{500}}\rfloor=\mathrm{22} \\ $$$${S}_{\mathrm{500}} =\mathrm{22}×\mathrm{501}−\frac{\mathrm{22}×\mathrm{23}×\mathrm{45}}{\mathrm{6}}=\mathrm{7227} \\ $$$$===================== \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\frac{{m}\left({m}+\mathrm{1}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${i}.{e}.\: \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{k}^{\mathrm{2}} \\ $$$${generally}: \\ $$$$\left[\sqrt[{{r}}]{\mathrm{1}}\right]+\left[\sqrt[{{r}}]{\mathrm{2}}\right]+\left[\sqrt[{{r}}]{\mathrm{3}}\right]+…+\left[\sqrt[{{r}}]{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{k}^{{r}} \\ $$$${with}\:{m}=\lfloor\sqrt[{{r}}]{{n}}\rfloor \\ $$
Commented by DAVONG last updated on 27/Mar/23
$$\mathrm{Thanks}\:\mathrm{sire} \\ $$