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\(=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{32}\right)+\left(\frac{1}{33}+...+\frac{1}{64}\right)\)
\(=1+\frac{1}{2}+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32\)
\(=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=1+\frac{1}{2}.6\)
\(=1+3\)
\(=4\)
~~ Bố thí cái li.ke ~~
Ta có :
A= 1+ 1/2 + 1/3 +1/4 + ...+ 1/63 + 1/64
=1 + ( 1/2 + 1/3 + 1/4 ) + ( 1/5 +1/6 + ..+1/8 ) + ( 1/9 + 1/10 + ..+ 1/16 ) + ( 1/17 + 1/18 + ...+ 1/32 ) + ( 1/33 + 1/34 + ...+1/63 + 1/64 )
=> A > 1 + ( 1/2 + 1/4.2 ) + 1/8.4 + 1/16.8 + 1/32.16 + 1/64.32
A > 1 + 1/2 + 1/2 + 1/2 +1/2
=>A > 4
Ta có :
\(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}=VP\left(đpcm\right)\)
Xét :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)
Thêm \(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\)vào mỗi vế ta có
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
\(\RightarrowĐPCM\)
Chứng minh rổng quát, Nếu:
\(A=\frac{1}{a^{2.k}}-\frac{1}{a^{2.\left(k+1\right)}}+\frac{1}{a^{2.\left(k+2\right)}}-\frac{1}{a^{2.\left(k+3\right)}}+...+\frac{1}{a^{2.\left(k+n\right)}}-\frac{1}{a^{2.\left(k+n+1\right)}}\) (a;b \(\in\) N*)
\(a^{2.k}.A=1-\frac{1}{a^{2.k}}+\frac{1}{a^{2.\left(k+1\right)}}-\frac{1}{a^{2.\left(k+2\right)}}+...+\frac{1}{a^{2.\left(k+n-1\right)}}-\frac{1}{a^{2.\left(k+n\right)}}\)
\(a^{2.k}.A+A=\left(1-\frac{1}{a^{2.k}}+\frac{1}{a^{2.\left(k+1\right)}}-\frac{1}{a^{2.\left(k+2\right)}}+..+\frac{1}{a^{2.\left(k+n-1\right)}}-\frac{1}{a^{2.\left(k+n\right)}}\right)-\left(\frac{1}{a^{2.k}}-\frac{1}{a^{2.\left(k+1\right)}}+\frac{1}{a^{2.\left(k+2\right)}}-\frac{1}{a^{2.\left(k+3\right)}}+..+\frac{1}{a^{2.\left(k+n\right)}}-\frac{1}{a^{2.\left(k+n+1\right)}}\right)\)
\(A.\left(a^{2.k}+1\right)=1-\frac{1}{a^{2.\left(k+n+1\right)}}< 1\)
\(A< \frac{1}{a^{2.k}+1}\)
Áp dụng vào bài toán dễ thấy a = 3; k = 1
Như vậy, \(A< \frac{1}{3^{2.1}+1}=\frac{1}{3^2+1}=\frac{1}{9+1}=\frac{1}{10}=0,1\left(đpcm\right)\)
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{2014}}-\frac{1}{3^{2016}}\)
\(\Rightarrow9A=1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{2012}}-\frac{1}{3^{2014}}\)
\(\Rightarrow10A=1-\frac{1}{3^{2016}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{2016}}}{10}\)
Vì 0,1 = \(\frac{1}{10}\) nên \(\frac{1-\frac{1}{3^{2016}}}{10}< \frac{1}{10}\) hay A < 0,1
1/2=1/2
1/3+1/4>1/4+1/4=1/2
1/5+…+1/8>4*1/8=1/2
1/9+…+1/16>8*1/16=1/2
1/2+1/3+1/4+…+1/16>4*1/2=2
1/2+1/3+1/4+…+1/63>1/2+1/3+1/4+…+1/16
=> 1/2+1/3+…+1/63>2
t i c k nhé !! 5756876876978080
Ta có:
\(\frac{1}{2}=\frac{1}{2}\)
\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+...+\frac{1}{8}>4.\frac{1}{8}=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)
\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}>4.\frac{1}{2}=2\)
\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}>\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}>2\)
a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(=\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^2\left(n^2+2n+1+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=>đpcm
b) Từ công thức trên ta có:
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=> \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)