Đáp án:
a, A = $\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^{10}}$
2A = 1 + $\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^9}$
2A - A = (1 + $\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^9}$) - $\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^{10}}$
⇒ A = 1 - $\frac{1}{2^{10}}$ = $\frac{1023}{1024}$
Vậy ...
b, B = 1 + $\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^{10}}$
Làm tương tự phần a ta được: 1 + $\frac{1023}{1024}$ = $\frac{2047}{1024}$
Vậy ...
c, C = 512 - $\frac{512}{2}$ - $\frac{512}{2^2}$ - ... - $\frac{512}{2^{10}}$
C = 512(1 - $\frac{1}{2}$ - $\frac{1}{2^2}$ - ... - $\frac{1}{2^{10}}$)
C = 512[1 - ($\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^{10}}$)]
Ta có: $\frac{1}{2}$ + $\frac{1}{2^2}$ + ... + $\frac{1}{2^{10}}$ = $\frac{1023}{1024}$ (Tính theo a)
⇒ C = 512(1 - $\frac{1023}{1024}$)
C = 512.$\frac{1}{1024}$ = $\frac{1}{2}$
Vậy ...
d, D = $\frac{1}{2}$ - $\frac{1}{2^2}$ + $\frac{1}{2^3}$ - ... + $\frac{1}{2^{99}}$ - $\frac{1}{2^{100}}$
2D = 1 - $\frac{1}{2}$ + $\frac{1}{2^2}$ - ... + $\frac{1}{2^{98}}$ - $\frac{1}{2^{99}}$
2D + D = (1 - $\frac{1}{2}$ + $\frac{1}{2^2}$ - ... + $\frac{1}{2^{98}}$ - $\frac{1}{2^{99}}$) + ($\frac{1}{2}$ - $\frac{1}{2^2}$ + $\frac{1}{2^3}$ - ... + $\frac{1}{2^{99}}$ - $\frac{1}{2^{100}}$)
3D = 1 - $\frac{1}{2^{100}}$
D = (1 - $\frac{1}{2^{100}}$) : 3 = $\frac{2^{100}-1}{3.2^{100}}$
Vậy ...
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