[Toán 11] Tính tổng

N

niemkieuloveahbu

Ta có:

[TEX]S_n=\frac{10-1}{9}+\frac{10^2-1}{9}+....+\frac{10^n-1}{9}\\= \frac{1}{9}(10+10^2+...++10^n)-\frac{n}{9}\\= \frac{1}{9}(10.\frac{10^n-1}{9})-\frac{n}{9}=\frac{10^{n+1}-10-9n}{81}[/TEX]
 
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