Question

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Answer 1

Salut,

Fracția de sub sumă nu depinde de j, deci avem că:

$\sum\limits_{j=1}^i\dfrac{i\cdot(i+1)}2=\dfrac{i\cdot(i+1)}2\cdot\sum\limits_{j=1}^i 1=\dfrac{i\cdot(i+1)}2\cdot i=\dfrac{i^2\cdot(i+1)}2\cdot$

Green eyes.