Chứng minh rằng:
\(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+ .... + \(\frac{1}{3^{99}}\)< \(\frac{1}{2}\)
a)Cho S = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2012!}.\) Chứng minh rằng S< 2
b)Chứng minh rằng :\(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+\frac{99}{100!}< \frac{1}{9!}\)
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sửa đề : \(\frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{99}{100!}\)
\(=\frac{10-1}{10!}+\frac{11-1}{11!}+\frac{12-1}{12!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=\frac{1}{9!}-\frac{1}{100!}< \frac{1}{9!}\left(đpcm\right)\)
Chứng minh rằng:
\(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+ .... + \(\frac{1}{3^{99}}\)< \(\frac{1}{2}\)
Lời giải:
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}(1)\)
\(\Rightarrow 3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}(2)\)
Lấy \((2)-(1)\Rightarrow 2A=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow A< \frac{1}{2}\)
Chứng minh rằng tổng \(P=\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\) nhỏ hơn 0, 1
\(P=\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\)
\(\Rightarrow9P=1-\frac{1}{3^2}+....+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\)
\(\Rightarrow9P+P=\left(1-\frac{1}{3^2}+....+\frac{1}{3^{2004}}-\frac{1}{3^{2006}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+....+\frac{1}{3^{2006}}-\frac{1}{3^{2008}}\right)\)
\(\Rightarrow10P=1-\frac{1}{3^{2008}}\)
\(\Rightarrow P=\frac{1}{10}-\frac{1}{3^{2008}\cdot10}< \frac{1}{10}=0,1\)
Vậy \(P< 0,1\)
Chứng minh rằng với mọi số tự nhiên n≥2
\(A=\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^2}+.....+\frac{1}{n^2}< \frac{2}{3}\)
Ta có \(\frac{1}{k^2}=\frac{4}{4k^2}< \frac{4}{4k^2-1}=2\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\left(k\in N\cdot\right)\)
Khi đó \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 2\left(\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\right)\\ =2\left(\frac{1}{3}-\frac{1}{2n+1}\right)< \frac{2}{3}\)
chứng minh rằng :\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.........+\frac{1}{100^2}< 1\)
Ta có:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A< 1-\frac{1}{100}< 1\left(đpcm\right)\)
giúp mk vs các bạn ưi ! mk đang cần gấp ai nhanh mik tích cho !nhanh nha help me!thank nhìu
Chứng minh rằng: \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}>2\)2
Chứng minh rằng :\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1\\ \)
Đặt A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2010^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{2010^2}< \frac{1}{2009\cdot2010}\)
=> A<\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)
\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2009}-\frac{2}{2010}\)
\(\Leftrightarrow A< 1-\frac{1}{2010}\)
<=> A<1 (đpcm)
Ta có \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{2010^2}< \frac{1}{2009.2010}\)
Cộng vế các BĐT trên ta được
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1-\frac{1}{2010}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}< 1\)
Câu 1: Cho \(a,b,c>0\)và \(a+b+c=3\). Chứng minh rằng:
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge\frac{3}{2}\).
Câu 2: Cho \(a,b,c,d>0\)và \(a+b+c+d=4\). Chứng minh rằng:
\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2}\ge2\).
Câu 3: Cho \(a,b,c,d>0\). Chứng minh rằng:
\(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+d^2}+\frac{d^3}{d^2+a^2}\ge\frac{a+b+c+d}{2}\).
Câu 4: Cho \(a,b,c,d>0\). Chứng minh rằng:
\(\frac{a^4}{a^3+2b^3}+\frac{b^4}{b^3+2c^3}+\frac{c^4}{c^3+2d^3}+\frac{d^4}{d^3+2a^3}\ge\frac{a+b+c+d}{3}\).
Câu 5: Cho \(a,b,c>0\)và \(a+b+c=3\). Chứng minh rằng:
\(\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\ge1\).
Câu 6: Cho \(a,b,c>0\)và \(a+b+c=3\). Chứng minh rằng:
\(\frac{a^2}{a+2b^3}+\frac{b^2}{b+2c^3}+\frac{c^2}{c+2a^3}\ge1\).
Câu 7: Cho \(a,b,c>0\)và \(a+b+c=3\). Chứng minh rằng:
\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge3\).
Câu 8: Cho \(a_1,a_2,...,a_{n-1},a_n>0\)và \(a_1+a_2+...+a_{n-1}+a_n=n\)với \(n\)nguyên dương. Chứng minh:
\(\frac{1}{a_1+1}+\frac{1}{a_2+1}+...+\frac{1}{a_{n-1}+1}+\frac{1}{a_n+1}\ge\frac{n}{2}\).
CHỨNG MINH RẰNG : \(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{4031}{2015^2.2016^2}< 1\)