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Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\) ta có :
\(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A>\frac{1}{2}-\frac{1}{2017}\)
\(A>\frac{2015}{4034}\) \(\left(1\right)\)
Lại có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A< 1-\frac{1}{2016}\)
\(A< \frac{2015}{2016}\) \(\left(2\right)\)
Từ (1) và (2) suy ra : \(\frac{2015}{4034}< A< \frac{2015}{2016}\) ( đpcm )
Vậy \(\frac{2015}{4034}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)
Chúc bạn học tốt ~
a, \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A< 1+\left(1-\frac{1}{100}\right)\Rightarrow A< 1+1-\frac{1}{100}\Rightarrow A< 2-\frac{1}{100}\Rightarrow A< 2\left(ĐPCM\right)\)
b, \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(\Rightarrow B< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2011\cdot2012}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\Rightarrow B< 1-\frac{1}{2012}\Rightarrow B< 1\left(1\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(\Rightarrow B>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2012\cdot2013}\)
\(\Rightarrow B>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(\Rightarrow B>\frac{1}{2}-\frac{1}{2013}\Rightarrow\frac{1}{2}-\frac{1}{2013}< B\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{2}-\frac{1}{2013}< B< 1\)
a)A=1+1/22+1/32+....+1/1002
<1+1/1.2+1/2.3+...+1/99.100=1+1-1/2+1/2-1/3+...+1/99-1/100=2-1/100=199/200<2
b)B=1/22+1/32+...+1/20122
<1/1.2+1/2.3+...+1/2011.2012=1-1/2+1/2-1/3+...+1/2011-1/2012=1-1/2012=2011/2012
1/2-1/2013=2011/4026<2011/2012<1
=> 2B= 2.(\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+...+\(\frac{1}{2^{2016}}\))
=>2B= \(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+...+\(\frac{1}{2^{2017}}\)
=>2B-B= \(\frac{1}{2^{2017}}\)- \(\frac{1}{2}\)
Mà \(\frac{1}{2}\) >\(\frac{1}{2^{2017}}\)
=>B<0<1 (đpcm)
\(\frac{B}{2}=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2017}}.\)
\(\frac{B}{2}=B-\frac{B}{2}=\frac{1}{2}-\frac{1}{2^{2017}}\Rightarrow B=1-\frac{1}{2^{2016}}< 1\)
Ta có \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\)<\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\)(đoạn này bn tự làm đc ko nếu ko thì thi nhắn cho mk) =\(1-\frac{1}{2016}\)
Do \(1-\frac{1}{2016}< 1\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< 1\)(đpcm)
Có 1/2^2+1/3^2+1/4^2+....+1/2016^2 <1/1.2+1/2.3+1/3.4+....+1/2015.2016(1)
Có 1/1.2+1/2.3+1/3.4+......+1/2015.2016
=1-1/2+1/2-1/3+1/3-1/4+........+1/2015-1/2016
=(-1/2+1/2)+(-1/3+1/3)+.........+(-1/2015+1/2015)+(1-1/2016)
=1-1/2016
=2016/2016-1/2016
=2015/2016(2)
Từ (1) và (2)
Suy ra 1/2^2+1/3^2+1/4^2+........+1/2016^2 <1
Đây là đpcm
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(< 1-\frac{1}{2016}< 1\left(đpcm\right)\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}< \frac{2015}{2016}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{2016}-\frac{1}{2017}\)
\(=\frac{1}{2}-\frac{1}{2017}=\frac{2015}{4024}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}>\frac{2015}{4034}\)
vậy ta có điều cần chứng minh
Lời giải:
$B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}$
$2B=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}$
Trừ theo vế:
$2B-B=1-\frac{1}{2^{2016}}$
$B=1-\frac{1}{2^{2016}}< 1$ (đpcm)