Chứng tỏ a/D = 1/101 + 1/102 + 1/103 + … +1/200 > 1/2 b/N = 1/2^2 + 1/3^2 + 1/4^2 + … + 1/100^2 < 1

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1. $D=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+…+\dfrac{1}{200}$

   Vì $\dfrac{1}{101};\dfrac{1}{102};…;\dfrac{1}{199}>\dfrac{1}{200}$

   $D>\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+…+\dfrac{1}{200}$

   $D>\dfrac{1}{200}.100$

   $D>\dfrac{100}{200}=\dfrac{1}{2}$

   $D>\dfrac{1}{2}$ $(đpcm)$

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2. Đáp án:

    

   Giải thích các bước giải:

   $a)$ $D=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+…+\dfrac{1}{200}$

   Ta có:

   $\dfrac{1}{101}$>$\dfrac{1}{200}$

   $ $

   $\dfrac{1}{102}$>$\dfrac{1}{200}$

   $\vdots$ 

   $\dfrac{1}{200}$=$\dfrac{1}{200}$

   $ $

   $⇒\dfrac{1}{101}+\dfrac{1}{102}+…+\dfrac{1}{200}$>$\dfrac{1}{200}+…+\dfrac{1}{200}$

   $ $

   $⇒\dfrac{1}{101}+\dfrac{1}{102}+…+\dfrac{1}{200}$>$\dfrac{100}{200}$

   $ $

   $⇒\dfrac{1}{101}+\dfrac{1}{102}+…+\dfrac{1}{200}$>$\dfrac{1}{2}$  (đpcm)

   $ $

   $ $

   $ $

   $b)$ $N=\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$

   Ta có:

   $\dfrac{1}{2^{2}}$<$\dfrac{1}{1.2}$

   $ $

   $\dfrac{1}{3^{2}}$<$\dfrac{1}{2.3}$

   $\vdots$ 

   $\dfrac{1}{100^{2}}$<$\dfrac{1}{99.100}$

   $ $ 

   $⇒\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$<$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{99.100}$

   $ $

   $⇒\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$<$1-\dfrac{1}{100}$

   $ $

   $⇒\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$<$\dfrac{99}{100}$

   $ $

   Mà $\dfrac{99}{100}$<$1$

   $ $

   $⇒\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+…+\dfrac{1}{100^{2}}$<$1$   (đpcm)

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