Question

Aratati ca:
$2(1+3+ 3^{2}+...+3^{8} ) \ \textless \ 3^{9}$

Answer 1

    
[tex]\displaystyle \\ \texttt{Vom folosi formula: } \\ \\ n^0+n^1+n^2+ \hdots + n^k = \frac{n^{k+1} - 1}{n-1} \\ \\ \texttt{Rezolvare: } \\ \\ 2(1+3+ 3^{2}+...+3^{8} ) =2 \times \frac{3^{8+1} -1}{3-1}= \\ \\ = \frac{2(3^{9} -1)}{2}= \boxed{ 3^{9} -1 \ \textless \ 3^9 } \\ \\ \texttt{cctd}[/tex]