[Toán 7] Đề thi Toán 7 học kì 1

T

tranduytrinh2000

Giải:
Ta có:[TEX] A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}[/TEX]
[TEX]A = \frac{2^{98} + 2^{96} + 2^{94} + ... + 1}{2^{100}}[/TEX] (quy đồng mẫu rồi cộng lại)
[TEX]2A = \frac{2^{99} + 2^{97} + 2^{95} + ... + 2}{2^{100}}[/TEX]
\Rightarrow [TEX]3A = \frac{2^{99} + 2^{98} + 2^{97} + 2^{96} + 2^{95} + ... + 1}{2^{100}}[/TEX]
[TEX]3A = \frac{2^{100} - 1}{2^{100}} = 1 - \frac{1}{2^{100}}[/TEX]
[TEX]A = \frac{1}{3}[/TEX] - [TEX]\frac{1}{2^{100} . 3} < \frac{1}{3}[/TEX]
:cool::cool::p
 
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0

0973573959thuy

Ta có: [TEX]A = \frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}[/TEX]

[TEX]A = \frac{2^{98} + 2^{96} + 2^{94} + ... + 1}{2^{100}}[/TEX]

[TEX]2A = \frac{2^{99} + 2^{97} + 2^{95} + ... + 2}{2^{100}}[/TEX]

[TEX]\rightarrow 3A = \frac{2^{99} + 2^{98} + 2^{97} + 2^{96} + 2^{95} + ... + 1}{2^{100}}[/TEX]

[TEX]3A = \frac{2^{100} - 1}{2^{100}} = 1 - \frac{1}{2^{100}}[/TEX]

[TEX]A = \frac{1}{3} - \frac{1}{2^{100} . 3} < \frac{1}{3}[/TEX] (dpcm)
 
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