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Question Number 11341 by Nadium last updated on 21/Mar/17

Commented by FilupS last updated on 22/Mar/17

Ω=lim_(n→∞) (1/n^4 )Σ_(i=1) ^n (Σ_(j=1) ^n (i^2 +j^2 )∣i^2 −j^2 ∣) i^2 −j^2 =(i+j)(i−j) ∴Σ_(j=1) ^n (i^2 +j^2 )∣i^2 −j^2 ∣=Σ_(j=1) ^n (i^2 +j^2 )∣(i+j)(i−j)∣ i, j >0 ∴=Σ_(j=1) ^n (i^2 +j^2 )(i+j)∣i−j∣ S=(i^2 +1)(i+1)∣i−1∣+(i^2 +2)(i+2)∣i−2∣+... ...+(i^2 +n)(i+n)∣i−n∣ Ω=lim_(n→∞) (1/n^4 )Σ_(i=1) ^n (Σ_(j=1) ^n (i^2 +j^2 )∣i^2 −j^2 ∣) Ω=lim_(n→∞) (1/n^4 )Σ_(i=1) ^n (S) Σ_(i=1) ^n (S)=(1^2 +1)(1+1)∣1−1∣+(1^2 +2)(1+2)∣1−2∣+... ...+(2^2 +1)(2+1)∣2−1∣+(2^2 +2)(2+2)∣2−2∣+... ...+(3^2 +1)(3+1)∣3−1∣+(3^2 +2)(3+2)∣3−2∣+... ......... =(2)(2)(0)+(3)(3)(1)+(4)(4)(2)+... ...+(5)(3)(1)+(6)(4)(0)+(7)(5)(1)+... ...+(10)(4)(2)+(11)(5)(1)+(12)(6)(0)+... ... =? working

$$\Omega=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)\mid{i}^{\mathrm{2}} −{j}^{\mathrm{2}} \mid\right) \\ $$$$\: \\ $$$${i}^{\mathrm{2}} −{j}^{\mathrm{2}} =\left({i}+{j}\right)\left({i}−{j}\right) \\ $$$$\therefore\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)\mid{i}^{\mathrm{2}} −{j}^{\mathrm{2}} \mid=\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)\mid\left({i}+{j}\right)\left({i}−{j}\right)\mid \\ $$$${i},\:{j}\:>\mathrm{0} \\ $$$$\therefore=\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)\left({i}+{j}\right)\mid{i}−{j}\mid \\ $$$${S}=\left({i}^{\mathrm{2}} +\mathrm{1}\right)\left({i}+\mathrm{1}\right)\mid{i}−\mathrm{1}\mid+\left({i}^{\mathrm{2}} +\mathrm{2}\right)\left({i}+\mathrm{2}\right)\mid{i}−\mathrm{2}\mid+... \\ $$$$\:\:\:\:\:...+\left({i}^{\mathrm{2}} +{n}\right)\left({i}+{n}\right)\mid{i}−{n}\mid \\ $$$$\: \\ $$$$\Omega=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)\mid{i}^{\mathrm{2}} −{j}^{\mathrm{2}} \mid\right) \\ $$$$\Omega=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({S}\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({S}\right)=\left(\mathrm{1}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)\mid\mathrm{1}−\mathrm{1}\mid+\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{1}+\mathrm{2}\right)\mid\mathrm{1}−\mathrm{2}\mid+... \\ $$$$\:\:\:\:\:\:\:\:...+\left(\mathrm{2}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}+\mathrm{1}\right)\mid\mathrm{2}−\mathrm{1}\mid+\left(\mathrm{2}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{2}+\mathrm{2}\right)\mid\mathrm{2}−\mathrm{2}\mid+... \\ $$$$\:\:\:\:\:\:\:\:...+\left(\mathrm{3}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}+\mathrm{1}\right)\mid\mathrm{3}−\mathrm{1}\mid+\left(\mathrm{3}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{3}+\mathrm{2}\right)\mid\mathrm{3}−\mathrm{2}\mid+... \\ $$$$......... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}\right)\left(\mathrm{2}\right)\left(\mathrm{0}\right)+\left(\mathrm{3}\right)\left(\mathrm{3}\right)\left(\mathrm{1}\right)+\left(\mathrm{4}\right)\left(\mathrm{4}\right)\left(\mathrm{2}\right)+... \\ $$$$\:\:\:\:\:\:\:\:\:...+\left(\mathrm{5}\right)\left(\mathrm{3}\right)\left(\mathrm{1}\right)+\left(\mathrm{6}\right)\left(\mathrm{4}\right)\left(\mathrm{0}\right)+\left(\mathrm{7}\right)\left(\mathrm{5}\right)\left(\mathrm{1}\right)+... \\ $$$$\:\:\:\:\:\:\:\:\:...+\left(\mathrm{10}\right)\left(\mathrm{4}\right)\left(\mathrm{2}\right)+\left(\mathrm{11}\right)\left(\mathrm{5}\right)\left(\mathrm{1}\right)+\left(\mathrm{12}\right)\left(\mathrm{6}\right)\left(\mathrm{0}\right)+... \\ $$$$... \\ $$$$=? \\ $$$${working} \\ $$

Commented by nume1114 last updated on 22/Mar/17

(i^2 +j^2 )∣i^2 −j^2 ∣ =∣i^2 +j^2 ∣∣i^2 −j^2 ∣ [i^2 +j^2 is always +ve] =∣(i^2 +j^2 )(i^2 −j^2 )∣ [∣a∣∣b∣=∣ab∣] =∣i^4 −j^4 ∣ if i>j then ∣i^4 −j^4 ∣=i^4 −j^4 i=j ∣i^4 −j^4 ∣=0 i<j ∣i^4 −j^4 ∣=j^4 −i^4 so, Σ_(i=1) ^n (Σ_(j=1) ^n ∣i^4 −j^4 ∣) = ∣1^4 −1^4 ∣+∣1^4 −2^4 ∣+∣1^4 −3^4 ∣+…+∣1^4 −n^4 ∣ +∣2^4 −1^4 ∣+∣2^4 −2^4 ∣+∣2^4 −3^4 ∣+…+∣2^4 −n^4 ∣ +∣3^4 −1^4 ∣+∣3^4 −2^4 ∣+∣3^4 −3^4 ∣+…+∣3^4 −n^4 ∣ ⋮ ⋱ +∣n^4 −1^4 ∣+∣n^4 −2^4 ∣+∣n^4 −3^4 ∣+…+∣n^4 −n^4 ∣ = (0)+(2^4 −1^4 )+(3^4 −1^4 )+…+(n^4 −1^4 ) +(2^4 −1^4 )+(0)+(3^4 −2^4 )+…+(n^4 −2^4 ) +(3^4 −1^4 )+(3^4 −1^4 )+(0)+…+(n^4 −2^4 ) ⋮ ⋱ +(n^4 −1^4 )+(n^4 −2^4 )+(n^4 −3^4 )+…+(0) = 2Σ_(n≥i>j≥1) (i^4 −j^4 ) =2Σ_(i=2) ^n (Σ_(j=1) ^(i−1) (i^4 −j^4 ))

$$\:\:\:\:\left({i}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)\mid{i}^{\mathrm{2}} −{j}^{\mathrm{2}} \mid \\ $$$$=\mid{i}^{\mathrm{2}} +{j}^{\mathrm{2}} \mid\mid{i}^{\mathrm{2}} −{j}^{\mathrm{2}} \mid\:\:\left[{i}^{\mathrm{2}} +{j}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{always}\:+\mathrm{ve}\right] \\ $$$$=\mid\left({i}^{\mathrm{2}} +{j}^{\mathrm{2}} \right)\left({i}^{\mathrm{2}} −{j}^{\mathrm{2}} \right)\mid\:\:\left[\mid{a}\mid\mid{b}\mid=\mid{ab}\mid\right] \\ $$$$=\mid{i}^{\mathrm{4}} −{j}^{\mathrm{4}} \mid \\ $$$$ \\ $$$$\mathrm{if}\:{i}>{j}\:\mathrm{then}\:\mid{i}^{\mathrm{4}} −{j}^{\mathrm{4}} \mid={i}^{\mathrm{4}} −{j}^{\mathrm{4}} \\ $$$$\:\:\:\:{i}={j}\:\:\:\:\:\:\:\:\:\:\:\mid{i}^{\mathrm{4}} −{j}^{\mathrm{4}} \mid=\mathrm{0} \\ $$$$\:\:\:\:{i}<{j}\:\:\:\:\:\:\:\:\:\:\:\mid{i}^{\mathrm{4}} −{j}^{\mathrm{4}} \mid={j}^{\mathrm{4}} −{i}^{\mathrm{4}} \\ $$$$\mathrm{so}, \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mid{i}^{\mathrm{4}} −{j}^{\mathrm{4}} \mid\right) \\ $$$$=\:\mid\mathrm{1}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \mid+\mid\mathrm{1}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \mid+\mid\mathrm{1}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \mid+\ldots+\mid\mathrm{1}^{\mathrm{4}} −{n}^{\mathrm{4}} \mid \\ $$$$\:+\mid\mathrm{2}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \mid+\mid\mathrm{2}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \mid+\mid\mathrm{2}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \mid+\ldots+\mid\mathrm{2}^{\mathrm{4}} −{n}^{\mathrm{4}} \mid \\ $$$$\:+\mid\mathrm{3}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \mid+\mid\mathrm{3}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \mid+\mid\mathrm{3}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \mid+\ldots+\mid\mathrm{3}^{\mathrm{4}} −{n}^{\mathrm{4}} \mid \\ $$$$\:\vdots\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ddots \\ $$$$\:+\mid{n}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \mid+\mid{n}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \mid+\mid{n}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \mid+\ldots+\mid{n}^{\mathrm{4}} −{n}^{\mathrm{4}} \mid \\ $$$$=\:\left(\mathrm{0}\right)+\left(\mathrm{2}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \right)+\left(\mathrm{3}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \right)+\ldots+\left({n}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \right) \\ $$$$\:+\left(\mathrm{2}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \right)+\left(\mathrm{0}\right)+\left(\mathrm{3}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \right)+\ldots+\left({n}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \right) \\ $$$$\:+\left(\mathrm{3}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \right)+\left(\mathrm{3}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \right)+\left(\mathrm{0}\right)+\ldots+\left({n}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \right) \\ $$$$\:\vdots\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ddots \\ $$$$\:+\left({n}^{\mathrm{4}} −\mathrm{1}^{\mathrm{4}} \right)+\left({n}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \right)+\left({n}^{\mathrm{4}} −\mathrm{3}^{\mathrm{4}} \right)+\ldots+\left(\mathrm{0}\right) \\ $$$$=\:\mathrm{2}\underset{{n}\geqslant{i}>{j}\geqslant\mathrm{1}} {\sum}\left({i}^{\mathrm{4}} −{j}^{\mathrm{4}} \right) \\ $$$$=\mathrm{2}\underset{{i}=\mathrm{2}} {\overset{{n}} {\sum}}\left(\underset{{j}=\mathrm{1}} {\overset{{i}−\mathrm{1}} {\sum}}\left({i}^{\mathrm{4}} −{j}^{\mathrm{4}} \right)\right) \\ $$

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