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Let [x] and x denote the greatest integers and the fractional parts of a real number x , and m belong to N then [x] + sigma x+m/100 (where m=1 to 100) is equal to

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Solution:    forall  min N,   x+m=x.

                   	herefore     [x]+sum_m=1^100 fracx+m100=[x]+frac100x100

                  Rightarrow             [x]+sum_m=1^100 fracx+m100=[x]+x=x.

              

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Deependra Verma

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