chứng minh rằng
a) 1/2^2 + 1/3^2 +...+ 1/10^2 < 1
b) 1/4^2 + 1/6^2 + 1/8^2 +...+ 1/(2n)^2 < 1/4
mình cần gấp
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\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
Cộng vế với vế ta được
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{99}{100}< 1\)
Vậy ta có đpcm
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}\)
\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};....;\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{49\cdot50}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< 1-\frac{1}{50}\)
\(\Rightarrow A< 1\Rightarrow1+A< 1+1=2\)
\(\Rightarrow\frac{1}{2^2}\cdot\left(1+A\right)< \frac{1}{2^2}\cdot2=\frac{1}{2}\)(đpcm)
Bài 1:
Ta có: \(2n^2\left(n+1\right)-2n\left(n^2+n-3\right)\)
\(=2n^3+2n^2-2n^3-2n^2+6n\)
\(=6n⋮6\)
1) \(2n^2\left(n+1\right)-2n\left(n^2+n-3\right)=2n^3+2n^2-2n^3-2n^2+6n=6n⋮6\forall n\in Z\)
2) \(n\left(3-2n\right)-\left(n-1\right)\left(1+4n\right)-1=3n-2n^2-4n^2+3n+1-1=-6n^2+6n=6\left(-n^2+n\right)⋮6\forall n\in Z\)
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2n-2\right).2n}\)
\(< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)
\(< \frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n}\right)=\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
\(\Rightarrow\) \(A< \frac{1}{4}\)
Study well ! >_<
a)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}< 1\)
Vậy \(A< 1\)
a) ta có: \(\frac{1}{2^2}\)<\(\frac{1}{1.2}\);\(\frac{1}{3^2}\)<\(\frac{1}{2.3}\)....\(\frac{1}{10^2}\)<\(\frac{1}{9.10}\)
Đặt A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{10^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{9.10}\)
\(\Rightarrow\)A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{9}\)-\(\frac{1}{10}\)
\(\Rightarrow\)A<1-\(\frac{1}{10}\)
\(\Rightarrow\)A<1