Question

A= 1+2+3+...+458, B= 15+16+17+...+52. Va rog sa ma ajutati.

Answer 1

[tex]\displaystyle \\ A= 1+2+3+...+458 = \frac{458(458+1)}{2}= \\ \\ = \frac{458\times 459)}{2}=229\times 459=\boxed{105111} \\ \\ \\ B=15+16+17+...+52 \\ \text{Daca sirul nu incepe de la 1, atunci se rezolva cu formula:} \\ \\ B = \frac{\text{(Nr. de termeni)(Ultimul termen + Primul termen)}}{2} \\ \\ \text{Calculam numarul de termeni: } \\ \\ n = 52 - 15 + 1 = 38 \text{ de termeni} \\ \\ B = \frac{38(52+15)}{2} = 19 \times 67 = \boxed{1273} [/tex]