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Question Number 73028 by mathmax by abdo last updated on 05/Nov/19
calculate Σ_(1≤i≤n and 1≤j≤n) min(i,j)
$${calculate}\:\sum_{\mathrm{1}\leqslant{i}\leqslant{n}\:{and}\:\mathrm{1}\leqslant{j}\leqslant{n}} \:\:{min}\left({i},{j}\right) \\ $$
Answered by mind is power last updated on 05/Nov/19
=Σ_(i=1) ^n i+2Σ_(i=2) ^n Σ_(j=1) ^(i−1) (j) =Σ_(i=1) ^n i+2Σ_(i=2) ^n .(((i−1).i)/2)=Σ_(i=1) ^n i−Σ_(i=2) ^n i+Σ_(i=2) ^n i^2 =1+Σ_(i=2) ^n i^2 =Σ_(i=1) ^n i^2 =((n(n+1)(2n+1))/6)
$$=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{i}+\mathrm{2}\underset{\mathrm{i}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\underset{\mathrm{j}=\mathrm{1}} {\overset{\mathrm{i}−\mathrm{1}} {\sum}}\left(\mathrm{j}\right) \\ $$$$=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{i}+\mathrm{2}\underset{\mathrm{i}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}.\frac{\left(\mathrm{i}−\mathrm{1}\right).\mathrm{i}}{\mathrm{2}}=\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{i}−\underset{\mathrm{i}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\mathrm{i}+\underset{\mathrm{i}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\mathrm{i}^{\mathrm{2}} =\mathrm{1}+\underset{\mathrm{i}=\mathrm{2}} {\overset{\mathrm{n}} {\sum}}\mathrm{i}^{\mathrm{2}} =\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{i}^{\mathrm{2}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 05/Nov/19
=2×[n×1+(n−1)×2+...+1×n]−(1+2+...+n) =2×Σ_(k=0) ^(n−1) (n−k)(k+1)−(1+2+...+n) =2×Σ_(k=0) ^(n−1) [(n−1)k+n−k^2 ]−(1+2+...+n) =2×[(n−1)((n(n−1))/2)+n^2 −(((n−1)n(2n−1))/6)]−((n(n+1))/2) =((n(n+1)(2n+1))/6)
$$=\mathrm{2}×\left[{n}×\mathrm{1}+\left({n}−\mathrm{1}\right)×\mathrm{2}+…+\mathrm{1}×{n}\right]−\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$$=\mathrm{2}×\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({n}−{k}\right)\left({k}+\mathrm{1}\right)−\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$$=\mathrm{2}×\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\left({n}−\mathrm{1}\right){k}+{n}−{k}^{\mathrm{2}} \right]−\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$$=\mathrm{2}×\left[\left({n}−\mathrm{1}\right)\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}+{n}^{\mathrm{2}} −\frac{\left({n}−\mathrm{1}\right){n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}\right]−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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