[Toán 7] Tính

B

braga

[TEX]1+2+3+4+....+3^{n+1}=\frac{3^{n+1}(3^{n+1}+1)}{2}[/TEX]

[TEX]\Rightarrow (1+2+3+4+....+3^{n+1}):2=\frac{3^{n+1}(3^{n+1}+1)}{2}:2=\frac{3^{n+1}(3^{n+1}+1)}{4}[/TEX]
 
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