InterviewSolution
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InterviewSolution
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If [x] denotes the greatest integer less than or equal to x then `lim_(n->oo)([x]+[2x]+[3x]+.....+[nx])/n^2`A. `x//2`B. `x//3`C. `x`D. 0 |
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Answer» Correct Answer - A For ant integer k, we have ` kx-1 lt [kx]le kx` `rArr Sigma_(k=1)^(n)(kx-1)lt Sigma_(k=1)^(n)[kx] le Sigma _(k=1)^(n)kx` `rArr (1)/(n^2)Sigma_(k=1)^(n)(kx-1)lt (1)/(n^2)Sigma_(k=1)^(n) [kx] le (1)/(n^2) Sigma _(k=1)^(n) kx` ` rArr (x)/(n^2)Sigma_(k=1)^(n) k-(1)/(n^2)lt (1)/(n^2) Sigma _(k=1)^(n) [kx] le (x)/(n^2)Sigma _(k=1)^(n) k` ` rArr (x)/(2) (1+(1)/(n)) -(1)/(n^2) lt (1)/(n^2) Sigma _(k=1) ^(n) [kx] le (x)/(2) (1+(2)/(n))` Now , ` lim_(xto oo) {(1+(1)/(n))-(1)/(n^2)}=(x)/(2) and, lim_(nto oo) (x)/(2) (1+(1)/(n))=(x)/(2)` ` therefore lim_(xto oo) (1)/(n^2)Sigma _(k=1)^(n)[kx]=(x)/(2)" "["Using Sandwich Theorem "]` i.e., `lim_(xtooo) ([x]+[2x]+[3x]+.....+[nx])/(n^2)=(x)/(2)` |
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