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\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-....+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
\(<\frac{1}{2^4}-\frac{1}{2^4}+\frac{1}{2^8}-\frac{1}{2^8}+...+\frac{1}{2^{4n}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2004}}-\frac{1}{2^{2004}}\)=0+0+0+...+0+....+0=0 <0,2
Vậy S<0,2
đặt A= \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)
B=\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{10000}{10001}\)
Lấy A.B= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10000}{10001}=\frac{1}{10001}\)
mặt khác
Ta có
\(\frac{1}{2}< \frac{2}{3}\\\)
\(\frac{3}{4}< \frac{4}{5}\)
....
\(\frac{9999}{10000}< \frac{10000}{10001}\)
=> A<B
=> A.A<A.B
=>A2<\(\frac{1}{10001}< \frac{1}{10000}\)
=>A<\(\sqrt{\frac{1}{10000}}=\frac{1}{100}\)
Vậy \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}\)<\(\frac{1}{100}\)
ĐPCM
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
\(3+\frac{1}{4+\frac{1}{b+\frac{1}{6}}}=\frac{421}{130}\) \(\Rightarrow\frac{1}{4+\frac{1}{b+\frac{1}{6}}}=\frac{31}{130}\Rightarrow4+\frac{1}{b+\frac{1}{6}}=\frac{130}{31}\Rightarrow\frac{1}{b+\frac{1}{6}}=\frac{6}{31}\Rightarrow b+\frac{1}{6}=\frac{31}{6}\Rightarrow b=\frac{30}{6}=5\)
Vậy b = 5
Đặt A = (1/2)(3/4)(5/6) ... (9999/10000) (A > 0)
.Và B = (2/3)(4/5)(6/7) ... (10000/10001) (B > 0)
Ta có A.B = (1/2)(2/3)(3/4) ... (10000/10001) = 1/10001 (1)
Mặt khác :
1/2 < 2/3
3/4 < 4/5
................
................
9999/10000 < 10000/10001
Nhân tất cả vế theo vế ---> A < B ---> A² < A.B (2)
(1),(2) ---> A² < 1/10001 ---> A < căn(1/10001) < căn(1/10000) = 1/100 (đpcm)
Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
và 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\)
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