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bài 2: Các số đó là :
-2012 , -2011 , -2010 , ....., 0, 1 , ..., 2012
Tổng cá số đó là 0
đúng nhé
Ta có A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^100
Suy ra 2A - A = ( 1 + 1/2 + 1/2^2 +...+ 1/2^99) - ( 1/2 + 1/2^2 +...+ 1/2^100 )
Suy ra A = 1 - 1/2^100 < 1
Vậy A < 1 ( ĐPCM)
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
Xét thấy : \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};...;\frac{1}{2013^2}< \frac{1}{2012\cdot2013}\)
Khi đó : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2012\cdot2013}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(=1-\frac{1}{2013}< 1\)
Hay \(A< 1\)
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}=\frac{2014}{2015}< 1\)
=> A < 1 (đpcm)
Nhanh cc ! ngu đừng hỏi lắm => càng hỏi càng ngu vvvv
Ta có : A= 1/2^2 +1/3^2 +....+1/2012^2 +1/2013^2
=> A= 1/2.2 +1/3.3 +....+1/2012.2012 +1/2013.2013
Do :1/2.2< 1/1.2
1/3.3 <1/2.3
.................
1/2012.2012 <1/2011.2012
1/2013.2013< 1/2012.2013
=>1/2.2 +1/3.3 +...+1/2012.2012+1/2013.2013< 1/1.2 +1/2.3+...+1/2011.2012+1/2012.2013
=>A<1/1 -1/2 +1/2 -1/3+...+1/2011-1/2012+1/2012-1/2013
=>A<1/1-1/2013
=>A<2013/2013 -1/2013
=> A< 2012/2013
Vì 2012<2013=>2012/2013<1
mà A<2012/2013=>A<1
Vậy A<1