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

Options

A. `x//2`
B. `x//3`
C. `x//6`
D. 0

Answer

B

Explanation

`k^2 x-1 lt [k^2x]le k^2x+1`
` rArr Sigma _(k=1)^(n) (k^2x-1)lt Sigma_(k=1)^(n) [k^2x]le Sigma_(k=3)^(n) Sigma _(k=1)^(n) (k^2x+1)`
` rArr (1)/(n^3) Sigma_(k=1)^(n) (k^2x-1)lt(1)/(n^3)Sigma_(k=1)^(n) [k^2x]le (1)/(n^3)Sigma _(k=1)^(n) (k^2x+1)`
` rArr (1)/(n^3) {(n(n+1)(2n+1))/(6)} x-(n)/(n^3)lt (1)/(n^3)Sigma _(k=1)^(n)[k^2x]le {(n(n+1)(2n+1))/(6)}x+(n)/(n^3)`
`rArr {(2n^2+3n+1)/(6n^2)}x-(1)/(n^2)lt (1)/(n^3)Sigma _(k=1)^(n) [k^2x]le {(2n^2+3n+1)/(6n^2)}x+(1)/(n^2)`
`therefore lim_(xto oo) (1)/(n^3)Sigma _(k=1)^(n) [k^2x]=(x)/(3)`