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myuyenfc605
2 năm trước
C= $\frac{1}{3}$ +$\frac{1}{3^2}$ +...+ $\frac{1}{3^99}$ $\frac{1}{3}$ +$\frac{1}{3^2}$ +...+ $\frac{1}{3^99}$ (phân số cuối là 3^99 nha) Chứng minh: C < $\frac{1}{2}$
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