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Đặt \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}\)là A
Ta có :A = \(\frac{1}{2}\left(1-\frac{1}{2^{100}}\right)\)
Vì 1-...
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.\(VT=\frac{c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{abc}{abc+bc+b}=\frac{c}{ac+c+1}+\frac{1}{ac+c+1}+\frac{ac}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1=VP\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(S=\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
\(\Rightarrow7^2S=1-\frac{1}{7^2}+\frac{1}{7^4}-\frac{1}{7^6}+....+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)
\(\Rightarrow49S=1-S-\frac{1}{7^{100}}\)
\(\Rightarrow49S+S=1-S-\frac{1}{7^{100}}+S\)
\(\Rightarrow50S=1-\frac{1}{7^{100}}<1\Rightarrow50S<1\Rightarrow S<\frac{1}{50}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giúp mình nha. Bài cuối cùng của đề toán dài 36 bài của mình đó
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)
Nên từ đây => \(A< 1\) (ĐPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}\)
\(B=\frac{1}{100}\)
TA co
22b=1+1/22+1/2^4+...+1/2^96+1/2^98
b=1/2^2+1/2^4+1/2^6+.......+1/2^98+1/2^100
tu 2 dong tren tru ve theo ve TA co 3b=1-1/200
suy ra b=1/1/200 /3=1/3-1/200 /3 be hon 1/3
nen b be hon 1/3
nghỉ hè rồi học cái gì nữa bạn ơi