[Toán 7]Toán Cm nè

Q

quynhnhung81

CMR
[TEX]\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + ...........+\frac{100}{3^{100}} < \frac{3}{4}[/TEX]
mời các bạn giải hộ :D:D:D:D:D

~~> Chú ý latex
[TEX]A=\frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + ...........+ \frac{100}{3^{100}} [/TEX]

[TEX]3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+ \frac {100}{3^{99}}[/TEX]

[TEX]2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+ \frac {1}{3^{99} }- \frac{100}{3^{100}}[/TEX]

[TEX]2A=\frac{1}{2}-\frac{1}{2.3^{99}} - \frac{100}{3^{100}} < \frac{1}{2}[/TEX]

[TEX]\Rightarrow A < \frac{1}{4} < \frac{3}{4}[/TEX]

.............................................................................................................................
 
S

soicon_boy_9x

Chị này làm sai rồi hay sao ý
Cách của mình nè:
[TEX]\blue A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{100}{3^{100}}[/TEX]

[TEX]\blue 3A=1+\frac{2}{3}+...+\frac{100}{3^{99}}[/TEX]

[TEX]\blue 3A-A=2A=1+\frac{1}{3}+...+\frac{1}{3^{99}-\frac{100}{3^{100}}[/TEX]

[TEX]\blue \rightarrow 6A=3+1+...+\frac{1}{3^{99}}-\frac{100}{3^{99}}[/TEX]

[TEX]\blue \rightarrow 6A-2A=4A=3-\frac{100}{3^{99}}+\frac{100}{3^{100}}[/TEX]
Bởi vì
[TEX]\blue \frac{100}{3^{99}}>\frac{100}{3^{100}}[/TEX]
nên $4A<3$
[TEX]\blue \rightarrow A<\frac{3}{4}[/TEX]
 
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