B=1\1×2×3×4+1\2×3×4×5+.....+1\21×22×23×24
So sánh B với 1\18
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Ta có: \(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}< \dfrac{1}{60}\times30\)
\(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}< \dfrac{1}{2}< \dfrac{3}{4}\)(đpcm)
Lời giải:
$\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}> \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}$
$\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}> \frac{10}{40}=\frac{1}{4}$
$\frac{1}{41}+\frac{1}{42}+....+\frac{1}{50}> \frac{10}{50}=\frac{1}{5}$
Cộng lại:
$A> \frac{1}{3}+\frac{1}{4}+\frac{1}{5}$
$A> \frac{47}{60}> \frac{3}{4}$
a. ( 23 - 21) + ( 19 - 17) + ( 15 - 13) + ( 11 - 9) + ( 7 - 5) + ( 3 - 1)
= 2 + 2 + 2 + 2 + 2 + 2
= 2 x 6
= 12
b. ( 24 - 22 ) + ( 20 - 18 ) + ( 16 - 14 ) + ( 12 - 10) + ( 8 - 6 ) + ( 4 - 2)
= 2 + 2 + 2 + 2 + 2 + 2
= 2 x 6
= 12
Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)