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prakhar raj ratna Grade: 12
        

lim n→∞  [x]+[2x]+[3x]+.........+[nx]




                                  n2


                                              

7 years ago

Answers : (2)

abhilash sainathan
14 Points
										

ans :x/2

7 years ago
anthony rebello
8 Points
										

I think the answer is                          X/2


 


heres how u do it :


 


[x] can be written as X+{x}    --> {x} is fractional part .. between 0 and 1


 


therefore ,

lim n→∞  [x]+[2x]+[3x]+.........+[nx]




                                  n2


is =

lim n→∞  x+2x+3x+.........+nx + {x}+{2x} ...+{nx}




                                  n2



= x(1+2+3+..+n)   +   {x}+...{nx}



                           n^2


 


=x(n(n+1))/2n^2 +  ({x}+...{nx})/n^2


 


Since {} is only b/w 0 and 1  , the second operand becomes 0 as n tends to


 


on solving the first part , u get (n2 x +nx)/2n2 = x/2 +x/2n


 


x/2n becomes 0 as x tends to infinity.


 


therefore the answer is x/2.


 


An easy way to do limits when Greatest integer  function is there, is to take out the commom element which does not change ... in this case x. (x , 2x, 3x ...)  x remains same.


divide this element by the power to which the denominator (having the dependent variable) is raised.


 


eg: [x]+[2x]+[3x]....+[nx]



                   n3


as n tends to infinity, = x/3




 


 


 


7 years ago
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