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# if a and b are +ve integers then find [a/b]+[2a/b]+[3a/b]......[(b-1)a/b] where [.] denotes greatest integer function

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we know    $\frac{a}{b}<=\left [ \frac{a}{b} \right ]<\frac{a}{b}+1$    (1)

let  U= [a/b]+[2a/b]+[3a/b]......[(b-1)a/b]

and applying the (1) in every term we get

$\left ( 1+2+...+b-1 \right )\frac{a}{b}<=U<\left ( 1+2+...+b-1 \right )\frac{a}{b}+\left ( b-1 \right )\times 1$

$=\frac{\left ( b \right )\left ( b-1 \right )}{2}\times \frac{a}{b}<= U < \frac{\left ( b \right )\left ( b-1 \right )}{2}\times \frac{a}{b} + \left ( b-1 \right )$

$=\frac{a\left ( b-1 \right )}{2}<=U<\frac{\left ( b-1 \right )\left ( a+2 \right )}{2}$

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