a)A= \frac{1}{3^1}311​+\frac{1}{3^2}321​+\frac{1}{3^3}331​+.........+\frac{1}{3^{99}}3991​

b)B=\frac{1}{3^1}311​+\frac{2}{3^2}322​+\frac{3}{3^3}333​+..........+\frac{99}{3^{99}}39999​

các bạn làm hộ mình nhé

a)A= \frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}​+.........+\frac{1}{3^{99}}

b)B=\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+..........+\frac{99}{3^{99}}

các bạn làm hộ mình nhé

a)A= \frac{1}{3^1}​+\frac{1}{3^2}​+\frac{1}{3^3}+.........+\frac{1}{3^{99}}

b)B=\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+..........+\frac{99}{3^{99}}

các bạn làm hộ mình nhé

a)A= \frac{1}{3^1}​+\frac{1}{3^2}​+\frac{1}{3^3}+.........+\frac{1}{3^{99}}

b)B=\frac{1}{3^1}+\frac{2}{3^2}+\frac{3}{3^3}+..........+\frac{99}{3^{99}}

các bạn làm hộ mình nhé

bài 1:chứng minh rằng:

a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)< \(\frac{1}{3}\)

b,\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)

Bài 2:

Bài vừa nãy là a. Bây giờ là câu b nè:

b)\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{99}+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)

a)A= \frac{1}{3^1}311​+\frac{1}{3^2}321​+\frac{1}{3^3}331​+.........+\frac{1}{3^{99}}3991​

b)B=\frac{1}{3^1}311​+\frac{2}{3^2}322​+\frac{3}{3^3}333​+..........+\frac{99}{3^{99}}39999​

các bạn làm hộ mình nhé

a)A= \frac{1}{3^1}311​+\frac{1}{3^2}321​+\frac{1}{3^3}331​+.........+\frac{1}{3^{99}}3991​

b)B=\frac{1}{3^1}311​+\frac{2}{3^2}322​+\frac{3}{3^3}333​+..........+\frac{99}{3^{99}}39999​

các bạn làm hộ mình nhé

Rút gọn:

\(B=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+.....+\frac{1}{3^{99}}\)