Question

3+3 la puterea2+.......3 la puterea x=120

Answer 1

   
[tex]\displaystyle\\ \text{Folosim formula:}\\\\ a^0+a^1+a^2+a^3+\cdots +a^n=\frac{a^{n+1} -1}{a-1},~~\text{unde }a \in N\\\\ \text{In enuntul dat lipseste }~3^0=1\\ \text{Din acest motiv, din formula de mai sus vom scadea }1=3^0\\\\ 3+3^2+3^3+\cdots 3^x=120\\\\ \frac{3^{x+1}-1}{3-1}-1=120\\\\ \frac{3^{x+1}-1}{2}=120+1\\\\ \frac{3^{x+1}-1}{2}=121\\\\ 3^{x+1}-1=121\cdot2\\ 3^{x+1}-1=242\\ 3^{x+1}=242+1\\ 3^{x+1}=243\\ \text{Observam ca }243=3^5\\ 3^{x+1}=3^5\\ x+1=5\\ x=5-1\\ \boxed{\bf x=4}[/tex]