Cho A=1/10+1/15+1/21+...+1/120
hãy chứng minh rằng A<1
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OY
27 tháng 7 2021
\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)
\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)
\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(M=2.\dfrac{3}{16}\)
\(M=\dfrac{3}{8}\)
Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)
P
0
NM
1
6 tháng 2 2018
Để chứng minh A<1/10 thì ta chứng minh A<2/3.4/5.6/7....100/101
Để chứng minh A>1/15 thì ta chứng minh A>1/2.2/3.4/5.98/99
5 tháng 4
a: Ta có
A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)
⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng
⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)
⇒ A > 1
vậy A > 1
b: ta có
S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))
⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))
⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)
⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)
⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)
vậy S > \(\dfrac{1}{2}\)
PY
0
TK
0
KQ
0
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)
Ta có :
\(\frac{1}{10}< 1\)
\(\frac{1}{15}< 1\)
\(\frac{1}{21}< 1\)
........................
\(\frac{1}{120}< 1\)
\(\Rightarrow\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}< 1\)
\(\Rightarrow A< 1\)( đpcm)
Ta có : A = \(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
= \(\frac{1}{20}\times2+\frac{1}{30}\times2+...+\frac{1}{240}\times2\)
= \(2\times\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
= \(2\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{15\times16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
= \(2\times\left(\frac{1}{4}-\frac{1}{16}\right)\)
= \(2\times\frac{3}{16}\)
= \(\frac{3}{8}\)< 1
=> A < 1