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Question Number 62878 by mathmax by abdo last updated on 26/Jun/19
calculate min Σ_(0≤i≤n and 0≤j≤n)     i.j
$${calculate}\:{min}\:\sum_{\mathrm{0}\leqslant{i}\leqslant{n}\:{and}\:\mathrm{0}\leqslant{j}\leqslant{n}} \:\:\:\:{i}.{j} \\ $$
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(n+1))/2)  =((n(n+1))/2).((n(n+1))/2)  =((n^2 (n+1)^2 )/4)
$$\sum_{\mathrm{0}\leqslant{i}\leqslant{n}\:{and}\:\mathrm{0}\leqslant{j}\leqslant{n}} \:\:\:\:{i}.{j} \\ $$$$=\sum_{\mathrm{0}\leqslant{i}\leqslant{n}} \sum_{\mathrm{0}\leqslant{j}\leqslant{n}} \:\:\:\:{i}.{j} \\ $$$$=\sum_{\mathrm{0}\leqslant{i}\leqslant{n}} {i}.\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}.\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by mathmax by abdo last updated on 26/Jun/19
thanks sir mrw.
$${thanks}\:{sir}\:{mrw}. \\ $$

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