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If a and b are positive integers with no common factor,show that `[a/b]+[(2a)/b]+[(3a)/b]. . . . . [((b-1)a)/b]=((a-1)(b-1))/2` when [.] shows the greatest integer function |
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Answer» Let `S = [a/b]+[(2a)/b]+[(3a)/b]+...+[((b-1)a)/b]->(1)` We can write it from end as, `S = [((b-1)a)/b]+ [((b-2)a)/b]+...[(2a)/b]+[a/b]->(2)` Adding (1) and (2), `2S = ([a/b]+[((b-1)a)/b])+([(2a)/b]+[((b-2)a)/b])+...+([((b-1)a)/b+[a/b]])` Now, `[a/b]+[((b-1)a)/b] = [a/b]+[a-a/b] = a-1` Similarly, `[(2a)/b]+[((b-2)a)/b] = [(2a)/b]+[a-(2a)/b] = a-1` So, each term will be equal to `a-1.` As, there are `b-1` terms, `:. 2S = (a-1)(b-1)` `=>S = ((a-1)(b-1))/2.` |
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