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
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
By Eliza
By Eliza
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
$D=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+…+\dfrac{1}{200}$
Vì $\dfrac{1}{101};\dfrac{1}{102};…;\dfrac{1}{199}>\dfrac{1}{200}$
$D>\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+…+\dfrac{1}{200}$
$D>\dfrac{1}{200}.100$
$D>\dfrac{100}{200}=\dfrac{1}{2}$
$D>\dfrac{1}{2}$ $(đpcm)$
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)