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Ta có A = 1/2+2/22+3/23+4/24+...+100/2100
<=> A = 1/2+2/4+3/9+4/16+...+100/2100
* Chứng minh \(\frac16
Ta có: \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)
\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)
\(\Rightarrow F<\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\cdots+\frac{1}{99\cdot100}\)
\(\) \(\Rightarrow F<\frac14-\frac15+\frac15-\frac16+\frac16-\frac17+\cdots+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow F<\frac14-\frac{1}{100}\)
\(\Rightarrow F<\frac{12}{25}\)
Mà \(\frac16=\frac{12}{72}<\frac{12}{25}\)
\(\Rightarrow\frac16 (1)
* Chứng minh \(F<\frac14\)
Lại có: \(\) \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)
\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)
\(\Rightarrow F>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\cdots+\frac{1}{100\cdot101}\)
\(\Rightarrow F>\frac15-\frac16+\frac16-\frac17+\frac17-\frac18+\cdots+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow F=\frac15-\frac{1}{101}\)
\(\Rightarrow F>\frac{96}{505}\)
Mà \(\frac14=\frac{96}{384}<\frac{96}{505}\)
\(\Rightarrow F<\frac14\) (2)
Từ (1) và (2) suy ra: \(\frac16
Vậy \(\frac16
Nếu \(n>0\Rightarrow\left(n-1\right)n\left(n+1\right)=n^3-n< n^3.\)
\(\Rightarrow VT< \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2005.2006.2007}\)
\(\Rightarrow2.VT< \frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2005.2006.2007}\)
\(\Rightarrow2.VT< \frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2007-2005}{2005.2006.2007}\)
\(\Rightarrow2VT< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\)
\(\Rightarrow2.VT< \frac{1}{2}-\frac{1}{2006.2007}\Rightarrow VT< \frac{1}{4}-\frac{1}{2.2006.2007}< \frac{1}{4}\)
M = 512 - 512/2 - .... - 512/2^10
= 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
= 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9 + 2^8 + 2^7 +... + 1 + 1/2
M = 2^10 - 2.2^9 + 1/2
M = 2^10 - 2^10 + 1/2
M = 1/2
Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
\(\Rightarrow49A=1-\frac{1}{7^2}+...+\frac{1}{7^{4n-4}}-\frac{1}{7^{4n}}+..+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)
\(\Rightarrow49A+A=50A=1-\frac{1}{7^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{50}=\frac{1}{50}-\frac{1}{7^{100}.50}< \frac{1}{50}\left(ĐPCM\right)\)
XET 1+1/2^2+1/3^2+1/4^2+....+1/100^2
=>1+1/2^2+1/3^2+1/4^2+....+1/100^2 < 1 + 1/1.2 + 1/2.3 +... + 1/100.101
=> 1+1/2^2+1/3^2+1/4^2+....+1/100^2 < 1+ (1-1/2+1/2-1/3+...+1/100-1/101)
=> 1+1/2^2+1/3^2+1/4^2+....+1/100^2 < 1+ 1-1/101 <2