a)Tính giá trị biểu thức A= 1/2 + 2/2^2 + 3/2^3 + 4/2^4 +... +10/2^10 B)Cho B=1/2^3 + 1/3^3+...+1/9^3 + 1/10^3

A) Ta có: A = $\frac{1}{2}$ + $\frac{2}{2^2}$ + $\frac{3}{2^3}$ + $\frac{4}{2^4}$ + ... + $\frac{10}{2^{10}}$ 2A = 1 + $\frac{2}{2}$ + $\frac{3}{2^2}$ + $\frac{4}{2^3}$ + ... + $\frac{10}{2^9}$ 2A - A = 1 + ($\frac{2}{2}$ - $\frac{1}{2}$) + ($\frac{3}{2^2}$ - $\frac{2}{2^2}$) + ... + ($\frac{10}{2^9}$ - $\frac{9}{2^9}$) - $\frac{10}{2^{10}}$ A = 1 + $\frac{1}{2}$ + $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + ... + $\frac{1}{2^9}$ - $\frac{10}{2^{10}}$ A = 1 + $\frac{\frac{1}{2}(1-(\frac{1}{2})^9)}{1-\frac{1}{2}}$ - $\frac{10}{2^{10}}$ A = 1 + $1 - (\frac{1}{2})^9$ - $\frac{10}{2^{10}}$ A = 2 - $(\frac{1}{2})^9$ - $\frac{10}{2^{10}}$ A = 2 - $\frac{1}{512}$ - $\frac{10}{1024}$ A = 2 - $\frac{1}{512}$ - $\frac{5}{512}$ A = 2 - $\frac{6}{512}$ A = 2 - $\frac{3}{256}$ A = $\frac{512}{256}$ - $\frac{3}{256}$ A = $\frac{509}{256}$ B) Ta có: B = $\frac{1}{2^3}$ + $\frac{1}{3^3}$ + ... + $\frac{1}{9^3}$ + $\frac{1}{10^3}$ Ta thấy rằng: $\frac{1}{2^3}$ < $\frac{1}{2^2}$, $\frac{1}{3^3}$ < $\frac{1}{3^2}$, ..., $\frac{1}{9^3}$ < $\frac{1}{9^2}$, $\frac{1}{10^3}$ < $\frac{1}{10^2}$ Do đó: B < $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... + $\frac{1}{9^2}$ + $\frac{1}{10^2}$ Mặt khác, ta lại có: $\frac{1}{2^2}$ < $\frac{1}{1×2}$, $\frac{1}{3^2}$ < $\frac{1}{2×3}$, ..., $\frac{1}{9^2}$ < $\frac{1}{8×9}$, $\frac{1}{10^2}$ < $\frac{1}{9×10}$ Do đó: B < $\frac{1}{1×2}$ + $\frac{1}{2×3}$ + ... + $\frac{1}{8×9}$ + $\frac{1}{9×10}$ B < 1 - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + ... + $\frac{1}{9}$ - $\frac{1}{10}$ B < 1 - $\frac{1}{10}$ B < $\frac{9}{10}$ Vậy B < $\frac{9}{10}$.