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If `[x]` denotes the greatest integer less than or equal to `x ,` then evaluate `("lim")_(ntooo)1/(n^2)([1.x]+[2.x]+[3.x]+....+[n.x])`

Text Solution

Verified by Experts

The correct Answer is:
`x//2`
`underset(ntooo)lim(1)/(n^(2))([1.x]+[2.x]+[3.x]+...+[n.x])`
`=underset(ntooo)lim{(sum_(r=1)^(n)[rx])/(n^(2))}`
`=underset(ntooo)lim{(sum_(r=1)^(n)(rx-{rx}))/(n^(2))}`
`=underset(ntooo)lim((x.(n(n+1))/(2))/(n^(2))-sum_(r=1)^(n)({rx})/(n^(2)))`
`=underset(ntooo)lim((x.(1+(1)/(n)))/(2)-sum_(r=1)^(n)({rx})/(n^(2)))`
`=x/2-0=x/2`
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