Question

Sa se determine numarul natural x pentru care 1+3+5+..+x=225

Answer 1

Termenii  sumei  sunt  in  progresie  aritmetica cu  ratia r=  2
Aplici  formula  sumei\
Sn=*a1+an)*n/2=(a1+a1+(n-1)r]/2=225  pt  ca  an=a1+(n-1)r
Faci  inlocuirile
[1+1+2(n-1)]*n/2=225  =>n²=225  n=15
x=an=1+(15-1)*2=29

Answer 2

$\displaystyle 1+3+5+...+x=225 ~~~~~~~~~~~~~~~~~~~~~r=2 \\ a_n=1+(n-1)\cdot 2=1+2n-2=2n-1 \\ S_n= \frac{a_1+a_n}{2} \cdot n \\ \frac{1+2n-1}{2} \cdot n=225 \\ \frac{2n}{2} \cdot n =225 \\ \frac{2n^2}{2} =225 \\ n^2=225 \\ n= \sqrt{225} \Rightarrow n=15 \in N \\ n=- \sqrt{225} \Rightarrow n=-15 \not \in N \\ a_n=a_1+(n-1) \cdot r=1+(15-1) \cdot 2=1+14 \cdot 2=1+28=29 \\ x=29$