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lim n→∞ [x]+[2x]+[3x]+.........+[nx]
n2
ans :x/2
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
is = lim n→∞ x+2x+3x+.........+nx + {x}+{2x} ...+{nx} n2
lim n→∞ x+2x+3x+.........+nx + {x}+{2x} ...+{nx}
= x(1+2+3+..+n) + {x}+...{nx}
=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
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