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Let [x] denotes greatest integer Then equal to
x/2
x/3
0
x/4
x/3 is ans
[x]=x-{x}
and
sum of sq. of n no. are n*(n+1)*(2n+1)/6
Dear Geetika Chouhan,
Ans:-Now the given function is a box function. So [x]=x-{x}
where {x}=fractional part of x whose value lie between 0 and 1
So writing all the box functions like this we get
the series (1²x+2²x+...........n²x) -K where K is a constant whose max value is n
Hence the sum is = xn(n+1)(2n+1)/6 -n
dividing it by n^3 we get
Sum=x(1+1/n)(2+1/n)/6 -1/n²
Taking n→∞ We get
Sum=x/3(ans)
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