CMR:1/2^2+1/3^2+1/4^2+...+1/100^2<1
Giúp mik vs ,ai trả lời trước mình tik cho
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\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
100=10*10
100=1000:10
100 câu nói hay về cuộc sống
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\Rightarrow A< 1\)
A=122+132+142+...+11002�=122+132+142+...+11002
A<11⋅2+12⋅3+13⋅4+...+199⋅100�<11⋅2+12⋅3+13⋅4+...+199⋅100
A<11−12+12−13+13−14+...+199−1100�<11−12+12−13+13−14+...+199−1100
A<1−1100�<1−1100
A<99100�<99100
Mà 99100<1⇒A<1
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\frac{1}{1\cdot2}< \frac{1}{2^2}\)
\(\frac{1}{2\cdot3}< \frac{1}{3^2}\)
\(\frac{1}{3\cdot4}< \frac{1}{4^2}\)
...
\(\frac{1}{99\cdot100}< \frac{1}{100^2}\)
\(\Rightarrow B< A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(B< A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{100}>\frac{3}{4}\)
\(\Leftrightarrow B< \frac{3}{4}< A\)
TA CÓ 1/2^2=1/4
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
1/100^2<1/99.100
=>1/2^2+2/3^2+.....+1/100^2<1/1.2+1/2.3+..+1/99.100
=1-99/100=99/100<1
Ta thấy:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}
Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy A<1
Học tốt nha!!!