4. 3a If a and b are positive integers with no common factor, show that f(b-1)a] (a -1)(b - 1) + ... + ., where [.) denotes the b function greatest integer

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( s=left[frac{a}{b}right]+left[frac{2 a}{b}right]+left[frac{3 n}{b}right]+ldots quadleft[frac{(b-1) a}{b}right] ) similarly let writing the terms in reverse order ( left[begin{array}{l}2 a_{0} bend{array}right]=I_{1}+f_{1} ) ( s=left[frac{(b-1) a}{b}right]+left[frac{(b-2) a}{b}right]+quadleft[frac{2 a}{b}right]+left[frac{a}{b}right] ) where ( I_{1} ) and ( f_{1} ) are the integral and fractional parts respectively ( 2 S=left(left[frac{a}{b}right]+left[frac{(b-1) a}{b}right]right)+left(left[frac{2 a}{b}right]+left[frac{(b-2) a}{b}right]right)+left(left[frac{(b-1) a}{b}right]+left[frac{a}{b}right]right) quadleft[frac{2 a}{b}right]+left[a-frac{2 a}{b}right]=left[I_{1}+f_{1}right]+left[a-I_{1}-f_{1}right] ) ( left.2 S=left(left[frac{a}{b}right]+left[a-frac{a}{b}right]right}+left(left[frac{2 a}{b}right]+left[a-frac{2 a}{b}right]right)+left(a-frac{a}{b}right]+left[frac{a}{b}right]right) quad=I_{1}+a-I_{1}-1 ) eq(3) result ( frac{a}{b}=I+f ) the integral part and f denotes the fracti ( 2 S=(a-1)(b-1) ) ( left[frac{a}{b}right]+left[a-frac{a}{b}right]=[1+f]+[a-I-f] ) ( =I+a-I-1 ) ( =a-1 )

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