Question Number 80804 by mr W last updated on 06/Feb/20
![Find [(√1)]+[(√2)]+[(√3)]+...+[(√(100))]=? with [x]=greatest integer function can we find a general formula for [(√1)]+[(√2)]+[(√3)]+...+[(√n)] in terms of n?](https://www.tinkutara.com/question/Q80804.png)
$${Find} \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{\mathrm{100}}\right]=? \\ $$$${with}\:\left[{x}\right]={greatest}\:{integer}\:{function} \\ $$$$ \\ $$$${can}\:{we}\:{find}\:{a}\:{general}\:{formula}\:{for}\: \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{{n}}\right] \\ $$$${in}\:{terms}\:{of}\:{n}? \\ $$
Commented by MJS last updated on 06/Feb/20
![[(√k)]=n; n^2 ≤k≤n(n+2) does this help?](https://www.tinkutara.com/question/Q80807.png)
$$\left[\sqrt{{k}}\right]={n};\:{n}^{\mathrm{2}} \leqslant{k}\leqslant{n}\left({n}+\mathrm{2}\right) \\ $$$$\mathrm{does}\:\mathrm{this}\:\mathrm{help}? \\ $$
Commented by mr W last updated on 07/Feb/20
$${yes},\:{it}\:{helps}.\:{thank}\:{you}\:{sir}! \\ $$
Answered by behi83417@gmail.com last updated on 07/Feb/20
![[(√1)]+[(√2)]+[(√3)]=3×1 [(√4)]+.........[(√8)]=5×2 [(√9)]+.........[(√(15))]=7×3 [(√(16))]+........[(√(24))]=9×4 [(√(25))]+........[(√(35))]=11×5 [(√(36))]+........[(√(48))]=13×6 [(√(49))]+.........[(√(63))]=15×7 [(√(64))]+.........[(√(80))]=17×8 [(√(81))]+.........[(√(100))]=19×9+10 Σ=3×1+5×2+7×3+..+19×9+10=625■](https://www.tinkutara.com/question/Q80824.png)
$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]=\mathrm{3}×\mathrm{1} \\ $$$$\left[\sqrt{\mathrm{4}}\right]+………\left[\sqrt{\mathrm{8}}\right]=\mathrm{5}×\mathrm{2} \\ $$$$\left[\sqrt{\mathrm{9}}\right]+………\left[\sqrt{\mathrm{15}}\right]=\mathrm{7}×\mathrm{3} \\ $$$$\left[\sqrt{\mathrm{16}}\right]+……..\left[\sqrt{\mathrm{24}}\right]=\mathrm{9}×\mathrm{4} \\ $$$$\left[\sqrt{\mathrm{25}}\right]+……..\left[\sqrt{\mathrm{35}}\right]=\mathrm{11}×\mathrm{5} \\ $$$$\left[\sqrt{\mathrm{36}}\right]+……..\left[\sqrt{\mathrm{48}}\right]=\mathrm{13}×\mathrm{6} \\ $$$$\left[\sqrt{\mathrm{49}}\right]+………\left[\sqrt{\mathrm{63}}\right]=\mathrm{15}×\mathrm{7} \\ $$$$\left[\sqrt{\mathrm{64}}\right]+………\left[\sqrt{\mathrm{80}}\right]=\mathrm{17}×\mathrm{8} \\ $$$$\left[\sqrt{\mathrm{81}}\right]+………\left[\sqrt{\mathrm{100}}\right]=\mathrm{19}×\mathrm{9}+\mathrm{10} \\ $$$$\Sigma=\mathrm{3}×\mathrm{1}+\mathrm{5}×\mathrm{2}+\mathrm{7}×\mathrm{3}+..+\mathrm{19}×\mathrm{9}+\mathrm{10}=\mathrm{625}\blacksquare \\ $$
Answered by mr W last updated on 07/Feb/20
![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=100: m=⌊(√(100))⌋=10 S_(100) =10×101−((10×11×21)/6)=625 with n=1000: m=⌊(√(1000))⌋=31 S_(1000) =31×1001−((31×32×63)/6)=20615 ===================== we see [(√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 this can be generalised: [(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) ⌋ e.g. [(1)^(1/5) ]+[(2)^(1/5) ]+[(3)^(1/5) ]+...+[(n)^(1/5) ]=m(n+1)−Σ_(k=1) ^m k^5 or [(1)^(1/5) ]+[(2)^(1/5) ]+[(3)^(1/5) ]+...+[(n)^(1/5) ]=m(n+1)−((m^2 (m+1)^2 (2m^2 +2m−1))/(12))](https://www.tinkutara.com/question/Q80835.png)
$${S}_{{n}} =\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{{n}}\right] \\ $$$${let}\:{m}=\lfloor\sqrt{{n}}\rfloor \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]=\mathrm{3}×\mathrm{1} \\ $$$$\left[\sqrt{\mathrm{4}}\right]+\left[\sqrt{\mathrm{5}}\right]+\left[\sqrt{\mathrm{6}}\right]+\left[\sqrt{\mathrm{7}}\right]+\left[\sqrt{\mathrm{8}}\right]=\mathrm{5}×\mathrm{2} \\ $$$$\left[\sqrt{\mathrm{9}}\right]+\left[\sqrt{\mathrm{10}}\right]+…+\left[\sqrt{\mathrm{15}}\right]=\mathrm{7}×\mathrm{3} \\ $$$$… \\ $$$$\left[\sqrt{{k}^{\mathrm{2}} }\right]+\left[\sqrt{{k}^{\mathrm{2}} +\mathrm{1}}\right]+…+\left[\sqrt{{k}^{\mathrm{2}} +\mathrm{2}{k}}\right]=\left(\mathrm{2}{k}+\mathrm{1}\right)×{k} \\ $$$$… \\ $$$$\left[\sqrt{{m}^{\mathrm{2}} }\right]+\left[\sqrt{{m}^{\mathrm{2}} +\mathrm{1}}\right]+…+\sqrt{{n}}=\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right)×{m} \\ $$$$ \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\sum}}\left(\mathrm{2}{k}+\mathrm{1}\right){k}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} =\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\sum}}{k}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\overset{{m}−\mathrm{1}} {\sum}}{k}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} =\mathrm{2}×\frac{\left({m}−\mathrm{1}\right){m}\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{6}}+\frac{\left({m}−\mathrm{1}\right){m}}{\mathrm{2}}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} =\frac{\left({m}−\mathrm{1}\right){m}\left(\mathrm{4}{m}+\mathrm{1}\right)}{\mathrm{6}}+\left({n}+\mathrm{1}−{m}^{\mathrm{2}} \right){m} \\ $$$${S}_{{n}} ={m}\left({n}+\mathrm{1}\right)−\frac{{m}\left({m}+\mathrm{1}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$ \\ $$$${with}\:{n}=\mathrm{100}: \\ $$$${m}=\lfloor\sqrt{\mathrm{100}}\rfloor=\mathrm{10} \\ $$$${S}_{\mathrm{100}} =\mathrm{10}×\mathrm{101}−\frac{\mathrm{10}×\mathrm{11}×\mathrm{21}}{\mathrm{6}}=\mathrm{625} \\ $$$$ \\ $$$${with}\:{n}=\mathrm{1000}: \\ $$$${m}=\lfloor\sqrt{\mathrm{1000}}\rfloor=\mathrm{31} \\ $$$${S}_{\mathrm{1000}} =\mathrm{31}×\mathrm{1001}−\frac{\mathrm{31}×\mathrm{32}×\mathrm{63}}{\mathrm{6}}=\mathrm{20615} \\ $$$$===================== \\ $$$${we}\:{see} \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\frac{{m}\left({m}+\mathrm{1}\right)\left(\mathrm{2}{m}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${i}.{e}.\: \\ $$$$\left[\sqrt{\mathrm{1}}\right]+\left[\sqrt{\mathrm{2}}\right]+\left[\sqrt{\mathrm{3}}\right]+…+\left[\sqrt{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{k}^{\mathrm{2}} \\ $$$${this}\:{can}\:{be}\:{generalised}: \\ $$$$\left[\sqrt[{{r}}]{\mathrm{1}}\right]+\left[\sqrt[{{r}}]{\mathrm{2}}\right]+\left[\sqrt[{{r}}]{\mathrm{3}}\right]+…+\left[\sqrt[{{r}}]{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{k}^{{r}} \\ $$$${with}\:{m}=\lfloor\sqrt[{{r}}]{{n}}\rfloor \\ $$$$ \\ $$$${e}.{g}. \\ $$$$\left[\sqrt[{\mathrm{5}}]{\mathrm{1}}\right]+\left[\sqrt[{\mathrm{5}}]{\mathrm{2}}\right]+\left[\sqrt[{\mathrm{5}}]{\mathrm{3}}\right]+…+\left[\sqrt[{\mathrm{5}}]{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{k}^{\mathrm{5}} \\ $$$${or} \\ $$$$\left[\sqrt[{\mathrm{5}}]{\mathrm{1}}\right]+\left[\sqrt[{\mathrm{5}}]{\mathrm{2}}\right]+\left[\sqrt[{\mathrm{5}}]{\mathrm{3}}\right]+…+\left[\sqrt[{\mathrm{5}}]{{n}}\right]={m}\left({n}+\mathrm{1}\right)−\frac{{m}^{\mathrm{2}} \left({m}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{12}} \\ $$
Commented by mr W last updated on 07/Feb/20
$${thank}\:{you}\:{for}\:{trying}\:{and}\:{reviewing}\:{sir}! \\ $$
Commented by behi83417@gmail.com last updated on 07/Feb/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{master}. \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{wonderful}\:\mathrm{and}\:\mathrm{great}. \\ $$