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đặt \(A=\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)
\(A=\left(\frac{2003}{2}+1\right)+\left(\frac{2002}{3}+1\right)+..+\left(\frac{1}{2004}+1\right)+\frac{2005}{2005}\)
\(A=\frac{2005}{2}+\frac{2005}{3}+..+\frac{2005}{2004}+\frac{2005}{2005}\)
\(A=2005.\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2004}+\frac{1}{2005}\right)\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2005}}{A}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2005}}{2005.\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2005}\right)}=\frac{1}{2005}\)
vậy P=1/2005
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\)
Suy ra: điều cần chứng minh
đặt 1/5^2+1/6^2+,,,+1/100^2=A
*chứng minh A<1/4
ta có: \(\frac{1}{5^2}=\frac{1}{5.5}<\frac{1}{4.5}\)
\(\frac{1}{6^2}=\frac{1}{6.6}<\frac{1}{5.6}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}<\frac{1}{99.100}\)
\(=>A<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A<\frac{1}{4}-\frac{1}{100}<\frac{1}{4}=>A<\frac{1}{4}\left(1\right)\)
*chứng minh A>1/6
ta có \(\frac{1}{5^2}=\frac{1}{5.5}>\frac{1}{5.6}\)
\(\frac{1}{6^2}=\frac{1}{6.6}>\frac{1}{6.7}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}>\frac{1}{100.101}\)
\(=>A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=>A>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}=>A>\frac{1}{6}\) (2)
từ (1) và (2)=>1/6<A<1/4 hay 1/6<1/5^2+...+1/100^2<1/4(đpcm)
tick nhé
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(A=\left(1-\frac{1}{\frac{\left(1+2\right).2}{2}}\right)\left(1-\frac{1}{\frac{\left(1+3\right).3}{2}}\right)...\left(1-\frac{1}{\frac{\left(1+2006\right).2006}{2}}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{2007.2006-2}{2006.2007}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}....\frac{2007.2006-2}{2006.2007}\) (1)
xét thấy:2007.2006-2=2006.(2008-1)+2006-2008=2006.(2008-1+1)-2008=2008.(2006-1)=2008.2005 (2)
(1),(2)\(=>A=\frac{4.1}{2.3}.\frac{5.2}{3.4}.\frac{6.3}{4.5}....\frac{2008.2005}{2006.2007}\)
\(A=\frac{\left(4.5.6...2008\right)\left(1.2.3...2005\right)}{\left(2.3.4....2006\right)\left(3.4.5...2007\right)}=\frac{2008}{2006.3}=\frac{1004}{3009}\)
Vậy A=1004/3009
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=11\)
\(\Leftrightarrow-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{n-1}+\sqrt{n}=11\)
\(\Leftrightarrow\sqrt{n}-1=11\Leftrightarrow\sqrt{n}=12\Leftrightarrow n=144\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{4}+\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Đặt \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(B=\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}\right)+\left(\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Giả sử tất cả các số hạng của B đều bằng \(\frac{1}{6^2}\)
\(\Rightarrow B=6.\frac{1}{6^2}=\frac{6}{36}=\frac{1}{6}<\frac{1}{4}\)
Do đó \(B<\frac{1}{4}\)
\(\Rightarrow A=\frac{1}{4}+B<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
Vậy \(A<\frac{1}{2}\)
Áp dụng công thức k/n*m=k/n-k/m trong đó n-m=k hoặc m-n=k
thế vào ta có
A=1/2*3+1/4*5+...+1/98*99
tớ biết tới đó thôi để từ từ tớ suy nghĩ rồi trả lời cho
A=1+1/2^2+1/3^2+1/4^2+...+1/100^2
A<1+1/1*2+1/2*3+1/3*4+...+1/99*100
A=1+1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1+1-1/100
A=2-1/100<2
nên A<2
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 2-\frac{1}{100}\)
Mà hiệu \(2-\frac{1}{100}< 2\Rightarrow A< 2\)
\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-....+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
\(<\frac{1}{2^4}-\frac{1}{2^4}+\frac{1}{2^8}-\frac{1}{2^8}+...+\frac{1}{2^{4n}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2004}}-\frac{1}{2^{2004}}\)=0+0+0+...+0+....+0=0 <0,2
Vậy S<0,2
Ảo quá \(\frac{1}{4n-2}<\frac{1}{4n}\)