công thức

N

nguyenbahiep1

[laTEX]S = 3^0 + 3^1+ ..+3^{n-1} \\ \\ 3.S = 3^1+3^2+..+3^{n-1}+3^{n} \\ \\ 3.S-S= 2S= 3^{n} - 3^0 = 3^n-1 \\ \\ S = \frac{3^n-1}{2} \\ \\ \Rightarrow 4.(3^0 + 3^1+ ..+3^{n-1} ) = 4.\frac{3^n-1}{2} = 2(3^n-1)[/laTEX]
 
Top Bottom