Formula De Gauss Nn+12

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$\begin{array}{l} \sum\limits_{k = 1}^n {\left( {2k - 1} \right)} = 2 \cdot \sum\limits_{k = 1}^n k - \sum\limits_{k = 1}^n 1 \\ \sum\limits_{k = 1}^n k = \frac{{n(n + 1)}}{2} \\ \sum\limits_{k = 1}^n 1 = n \\ - - - - - - - - \\ \sum\limits_{k = 1}^n {\left( {2k - 1} \right)} = 2 \cdot \sum\limits_{k = 1}^n k - \sum\limits_{k = 1}^n 1 = 2 \cdot \frac{{n(n + 1)}}{2} - n = n^2 \\ {\rm Pt exercitiul tau,care cere sa se calculeze :} \\ {\rm 1 + 3 + 5 + }...{\rm + 99 = ??} \\ {\rm Putem fie folosim rezultatul de mai sus(mai scurt) fi ne folosim de observatia:} \\ {\rm 3 = 2 + 1} \\ {\rm 5 = 4 + 1} \\ {\rm 7 = 6 + 1} \\ .......... \\ {\rm 99 = 98 + 1} \\ {\rm Adunand toate sumele astea seobtine:} \\ {\rm 1 + 3 + }...{\rm + 99 = 1 + 2 + 1 + 4 + 1 + }....{\rm + 98 + 1} \\ {\rm = 50} \cdot {\rm 1 + 2 + }...{\rm + 98} \\ {\rm = 50 + 2(}\underbrace {{\rm 1 + }...{\rm + 49}}_{{\rm Suma}{\rm .Gauss}}{\rm )} \\ {\rm = 50 + 2}\left( {\frac{{{\rm 49} \cdot {\rm 50}}}{{\rm 2}}} \right) = ... = 50^2 \\ \end{array}$

Calculati 1 3 5 99 Si 2 4 6 100 Forum Anideșcoală Ro