Menu Close

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-




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?
Find[1]+[2]+[3]++[100]=?with[x]=greatestintegerfunctioncanwefindageneralformulafor[1]+[2]+[3]++[n]intermsofn?
Commented by MJS last updated on 06/Feb/20
[(√k)]=n; n^2 ≤k≤n(n+2)  does this help?
[k]=n;n2kn(n+2)doesthishelp?
Commented by mr W last updated on 07/Feb/20
yes, it helps. thank you sir!
yes,ithelps.thankyousir!
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■
[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+10Σ=3×1+5×2+7×3+..+19×9+10=625◼
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))
Sn=[1]+[2]+[3]++[n]letm=n[1]+[2]+[3]=3×1[4]+[5]+[6]+[7]+[8]=5×2[9]+[10]++[15]=7×3[k2]+[k2+1]++[k2+2k]=(2k+1)×k[m2]+[m2+1]++n=(n+1m2)×mSn=m1k=1(2k+1)k+(n+1m2)mSn=2m1k=1k2+m1k=1k+(n+1m2)mSn=2×(m1)m(2m1)6+(m1)m2+(n+1m2)mSn=(m1)m(4m+1)6+(n+1m2)mSn=m(n+1)m(m+1)(2m+1)6withn=100:m=100=10S100=10×10110×11×216=625withn=1000:m=1000=31S1000=31×100131×32×636=20615=====================wesee[1]+[2]+[3]++[n]=m(n+1)m(m+1)(2m+1)6i.e.[1]+[2]+[3]++[n]=m(n+1)mk=1k2thiscanbegeneralised:[1r]+[2r]+[3r]++[nr]=m(n+1)mk=1krwithm=nre.g.[15]+[25]+[35]++[n5]=m(n+1)mk=1k5or[15]+[25]+[35]++[n5]=m(n+1)m2(m+1)2(2m2+2m1)12
Commented by mr W last updated on 07/Feb/20
thank you for trying and reviewing sir!
thankyoufortryingandreviewingsir!
Commented by behi83417@gmail.com last updated on 07/Feb/20
thank you very much dear master.  it is wonderful and great.
thankyouverymuchdearmaster.itiswonderfulandgreat.

Leave a Reply

Your email address will not be published. Required fields are marked *