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Georgianaak
a fost răspuns

Rezolvati :Suma de la k=1 la n din K(k+2). Multumesc!

Răspuns :

[tex]\displaystyle \sum\limits_{k=1}^nk(k+2)=\sum\limits_{k=1}^n(k^2+2k)=\sum\limits_{k=1}^nk^2+\sum\limits_{k=1}^n2k=\sum\limits_{k=1}^nk^2+2\sum\limits_{k=1}^nk=\\ \\ = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{ 2} = \frac{n(n+1)}{2} \left( \frac{2n+1}{3} +2\right)= \\ \\ = \frac{n(n+1)}{2} \cdot \frac{2n+1+6}{3} = \frac{n(n+1)}{2} \cdot \frac{2n+7}{3} = \frac{n(n+1)(2n+7)}{6} [/tex]

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