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\(S=\frac{1}{1.1.3}+\frac{1}{2.3.5}+\frac{1}{3.5.7}+\frac{1}{4.7.9}+...+\frac{1}{100.199.201}\)
\(S=\frac{1}{3}+\frac{2}{4.3.5}+\frac{2}{6.5.7}+\frac{2}{8.7.9}+...+\frac{2}{200.199.201}\)
Ta có: \(\frac{2}{3.4.5}< \frac{2}{3.5}\)
\(\Rightarrow S< \frac{1}{3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{199.201}\)
\(\Rightarrow S< \frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{199}-\frac{1}{201}\)
\(\Rightarrow S< \frac{1}{3}+\frac{1}{3}-\frac{1}{201}\)
\(\Rightarrow S< \frac{2}{3}-\frac{1}{201}< \frac{2}{3}\)
\(\Rightarrow S< \frac{2}{3}\)
Chúc học tốt.
Lời giải:
a) Số hạng thứ $n$: \(\frac{1}{n(2n-1)(2n+1)}\)
b) Tổng $A$ có 2011 số hạng có dạng là:
\(A=\frac{1}{1.1.3}+\frac{1}{2.3.5}+....+\frac{1}{2011.4021.4023}\)
\(A=\frac{2}{2.1.3}+\frac{2}{4.3.5}+\frac{2}{6.5.7}+....+\frac{2}{4022.4021.4023}\)
\(=\frac{2}{1.2.3}+\frac{2}{3.4.5}+\frac{2}{5.6.7}+...+\frac{2}{4021.4022.4023}\)
\(< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2011.2012.2013}\)
$A< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2013-2011}{2011.2012.2013}$
$A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-....-\frac{1}{2012.2013}$
$A< \frac{1}{2}-\frac{1}{2012.2013}< \frac{1}{2}< \frac{2}{3}$
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..............
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)
ĐPcm
Giải:
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\)
\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\)
\(\Rightarrow S< 1\)
Vậy S < 1.
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{2}{5}\)
ô cái đầu là \(\frac{1}{1.1.2}\)muk
SAo không ai giúp thế này?