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)
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)
Commented by mr W last updated on 07/Feb/20

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)
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)
Commented by mr W last updated on 07/Feb/20

Commented by behi83417@gmail.com last updated on 07/Feb/20
