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

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

$$\underset{k=1}{\sum ^{100}}\mid i-k\mid$$$$=\underset{k=1}{\sum ^i}\mid i-k\mid +\underset{k=i+1}{\sum ^{100}}\mid i-k\mid$$$$=\underset{k=1}{\sum ^i}(i-k)+\underset{k=i+1}{\sum ^{100}}(k-i)$$$$=\frac{(i-1)i}{2}+\frac{(100-i)(101-i)}{2}$$$$=\frac{2i^2+10100-202i}{2}$$$$=i^2-101i+5050$$$$\underset{i=1}{\sum ^x}\underset{k=1}{\sum ^{100}}\mid i-k\mid$$$$=\underset{i=1}{\sum ^x}(i^2-101i+5050)$$$$=\frac{x(x+1)(2x+1)}{6}-\frac{101x(x+1)}{2}+5050x$$$$=\frac{x(x^2-150x+14999)}{3}=333300$$$$\Rightarrow x=100✓$$

Commented by hardmath last updated on 12/Jan/24

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

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