Toán nâng cao 7

M

Minions1234

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Q

quynhphamdq

Câu 2 :
Ta có : $B=\frac{1}{2}+(\frac{1}{2})^2 + ...+ (\frac{1}{2})^{98} + (\frac{1}{2})^{99}$
\Rightarrow$\frac{1}{2}B= (\frac{1}{2})^2 + ...+ (\frac{1}{2})^{99} + (\frac{1}{2})^{100}$
\Rightarrow $2B-B=[(\frac{1}{2})^2 + ...+ (\frac{1}{2})^{99} + (\frac{1}{2})^{100}]-[\frac{1}{2}+(\frac{1}{2})^2 + ...+ (\frac{1}{2})^{98} + (\frac{1}{2})^{99}]$
\Rightarrow $B=1-(\frac{1}{2})^{100}$ < 1 (đpcm)
 
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