Question-203206

Question Number 203206 by hardmath last updated on 12/Jan/24

Answered by mr W last updated on 12/Jan/24

Σ_(k=1) ^(100) ∣i−k∣ =Σ_(k=1) ^i ∣i−k∣+Σ_(k=i+1) ^(100) ∣i−k∣ =Σ_(k=1) ^i (i−k)+Σ_(k=i+1) ^(100) (k−i) =(((i−1)i)/2)+(((100−i)(101−i))/2) =((2i^2 +10100−202i)/2) =i^2 −101i+5050 Σ_(i=1) ^x Σ_(k=1) ^(100) ∣i−k∣ =Σ_(i=1) ^x (i^2 −101i+5050) =((x(x+1)(2x+1))/6)−((101x(x+1))/2)+5050x =((x(x^2 −150x+14999))/3)=333300 ⇒x=100 ✓

$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\mid{i}−{k}\mid \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}\mid{i}−{k}\mid+\underset{{k}={i}+\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\mid{i}−{k}\mid \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}\left({i}−{k}\right)+\underset{{k}={i}+\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\left({k}−{i}\right) \\ $$$$=\frac{\left({i}−\mathrm{1}\right){i}}{\mathrm{2}}+\frac{\left(\mathrm{100}−{i}\right)\left(\mathrm{101}−{i}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{i}^{\mathrm{2}} +\mathrm{10100}−\mathrm{202}{i}}{\mathrm{2}} \\ $$$$={i}^{\mathrm{2}} −\mathrm{101}{i}+\mathrm{5050} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{x}} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{100}} {\sum}}\mid{i}−{k}\mid \\ $$$$=\underset{{i}=\mathrm{1}} {\overset{{x}} {\sum}}\left({i}^{\mathrm{2}} −\mathrm{101}{i}+\mathrm{5050}\right) \\ $$$$=\frac{{x}\left({x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}{\mathrm{6}}−\frac{\mathrm{101}{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}+\mathrm{5050}{x} \\ $$$$=\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{150}{x}+\mathrm{14999}\right)}{\mathrm{3}}=\mathrm{333300} \\ $$$$\Rightarrow{x}=\mathrm{100}\:\checkmark \\ $$

Commented by hardmath last updated on 12/Jan/24

perfect my dear professor thank you so much

$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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