Question

Daca 1+3+5+...+13=x la puterea 2 .
Atunci x este : a)7
b)26
c)49
d)6

Answer 1

$\displaystyle 1+3+5+...+13=x^2 \\ \\ 1+2+3+4+5+...+13-(2+4+6+...+12)=x^2 \\ \\ \frac{13(13+1)}{2} -2(1+2+3+...+6)=x^2 \\ \\ \frac{13 \cdot 14}{2} -2 \cdot \frac{6(6+1)}{2} =x^2 \\ \\ \frac{182}{2} -2 \cdot \frac{6 \cdot 7}{2} =x^2 \\ \\ 91-\not 2 \cdot \frac{42}{\not 2}=x^2 \\ \\ 91-42=x^2 \\ \\ 49=x^2 \\ \\ x= \sqrt{49} \\ \\ \boxed{x=7}$