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Si0140896
a fost răspuns

E(x)=(2x - 1)^2 - 3(x - 3)(x+2) - (x - 2)(x + 1 )

Demostrati ca E(1)+E(1/2)+E(1/3)+ ... +E(1/2020)=42420

Răspuns :

[tex]\displaystyle\\E(x)=(2x-1)^2-3(x-3)(x+2)-(x-2)(x+1 )=\\4x^2-4x+1-3(x^2+2x-3x-6)-(x^2+x-2x-2)=\\4x^2-4x+1-3(x^2-x-6)-(x^2-x-2)=\\4x^2-4x+1-3x^2+3x+18-x^2+x+2=18+1+2=21.\\Asadar,~\sum_{i=1}^{2020}E\left(\frac{1}i\right)=21\cdot2020=42420[/tex]

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