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Question Number 62879 by mathmax by abdo last updated on 26/Jun/19

calculate min Σ_(0≤i≤n  and 0≤j≤n)  (i+j)

$${calculate}\:{min}\:\sum_{\mathrm{0}\leqslant{i}\leqslant{n}\:\:{and}\:\mathrm{0}\leqslant{j}\leqslant{n}} \:\left({i}+{j}\right) \\ $$

Answered by mr W last updated on 26/Jun/19

Σ_(0≤i≤n  and 0≤j≤n)  (i+j)  =Σ_(0≤i≤n ) Σ_(0≤j≤n)  (i+j)  =Σ_(0≤i≤n )  (i+((n(1+n))/2))  =((n(n+1))/2)+((n(1+n))/2)  =n(n+1)

$$\sum_{\mathrm{0}\leqslant{i}\leqslant{n}\:\:{and}\:\mathrm{0}\leqslant{j}\leqslant{n}} \:\left({i}+{j}\right) \\ $$$$=\sum_{\mathrm{0}\leqslant{i}\leqslant{n}\:} \sum_{\mathrm{0}\leqslant{j}\leqslant{n}} \:\left({i}+{j}\right) \\ $$$$=\sum_{\mathrm{0}\leqslant{i}\leqslant{n}\:} \:\left({i}+\frac{{n}\left(\mathrm{1}+{n}\right)}{\mathrm{2}}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\frac{{n}\left(\mathrm{1}+{n}\right)}{\mathrm{2}} \\ $$$$={n}\left({n}+\mathrm{1}\right) \\ $$

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