Question

rút các biểu thức sau

1+1/3+1/3^2+1/3^3+....+1/3^100

GIải đầy đủ hộ mình nhé

o l m . v n

Answer 1

    Đặt        \(A=1+\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{100}\)

                 \(\frac{1}{3}A=\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+.....+\frac{1}{3}^{101}\)

          \(\frac{1}{3}A-A=\left(\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{101}\right)-\left(1+\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{100}\right)\)

          \(\frac{1}{3}A-A=\frac{1}{3}+\frac{1}{3}^2+\frac{1}{3}^3+....+\frac{1}{3}^{101}-1-\frac{1}{3}-\frac{1}{3}^2-\frac{1}{3}^3-....-\frac{1}{3}^{100}\)

          \(\frac{\left(-2\right)}{3}A=\frac{1}{3}-1\)

          \(\frac{\left(-2\right)}{3}A=\frac{\left(-2\right)}{3}\Rightarrow A=1\)

Vậy ......