Chứng tỏ 1/1.2+1/1.2.3+1/1.2.3.4+..+1/1.2.3.4.5…100<1

Đáp án:

Giải thích các bước giải:

 $\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+\dfrac{1}{1.2.3.4}+…+\dfrac{1}{1.2.3….100}$

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Ta có: $\dfrac{1}{1.2}=\dfrac{1}{1.2}$ ; $\dfrac{1}{1.2.3}=\dfrac{1}{2.3}$ ; $\dfrac{1}{1.2.3.4}<\dfrac{1}{3.4}$ ; … ; $\dfrac{1}{1.2.3…..100}<\dfrac{1}{99.100}$

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$⇒\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+\dfrac{1}{1.2.3.4}+…+\dfrac{1}{1.2.3….100}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{99.100}$

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$⇒\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+\dfrac{1}{1.2.3.4}+…+\dfrac{1}{1.2.3….100}<1-\dfrac{1}{100}$

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$⇒\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+\dfrac{1}{1.2.3.4}+…+\dfrac{1}{1.2.3….100}<\dfrac{99}{100}$

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$⇒\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+\dfrac{1}{1.2.3.4}+…+\dfrac{1}{1.2.3….100}<\dfrac{99}{100}<1$

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$⇒\dfrac{1}{1.2}+\dfrac{1}{1.2.3}+\dfrac{1}{1.2.3.4}+…+\dfrac{1}{1.2.3….100}<1$      (đpcm)