cm: 1/2^2+1/3^2+1/4^2+...+1/100^2< 1
Những câu hỏi liên quan
C=1/2^2+1/3^2+1/4^2+...+1/19^2+1/20^2
CM C <3/4
D=1/2^2+1/3^2+1/4^2+...+1/100^2
CM D<1
cm 1/2^2+1/3^2+1/4^2+...+1/100^2<1/3/4
A=1/2^2 +1/2^4 +1/2^6 +1/2^8 +⋯+1/2^100 CM: A<1/3
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow4A=2^2\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}\right)=1+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}\)
\(\Rightarrow3A=4A-A=1+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^2}-\dfrac{1}{2^4}-...-\dfrac{1}{2^{100}}=1-\dfrac{1}{2^{100}}\)
\(\Rightarrow A=\left(1-\dfrac{1}{2^{100}}\right):3=\dfrac{1}{3}-\dfrac{1}{2^{100}.3}< \dfrac{1}{3}\left(đpcm\right)\)
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Cm : A = 1/1^2+ 1/2^2 + 1/3^2+ ...... +1/99^2+1/100^2 <7/4
A=1/1^2+ 1/2^2+ 1/3^2+...+ 1/99^2+ 1/100^2
A=1+ 1/2^2+ 1/3^2+...+ 1/99^2+ 1/100^2
A<1+(1/2^2+1/2.3+1/3/4+...+1/98.99+1/99.100) (giữ nguyên phân số 1/2^2)
A<1+ (1/4+1/2-1/3+1/3-1/4+...+1/99-1/99+1/99-1/100)
A<1+ (1/4+1/2-1/100)
Mà 1/4+1/2-1/100 <1/4+1/2=3/4
=>A<1+3/4=7/4
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1 + 1 + x = 3
2 + x = 3
x = 3 - 2
x = 1
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CM rằng : A = 1+1/3^2+1/4^2+…+1/100^2 < 1
cứuuuuu 🫠
Sửa đề: A=1/2^2+...+1/100^2
1/2^2<1/1*2
1/3^2<1/2*3
...
1/100^2<1/99*100
=>A<1-1/2+1/2-1/3+...+1/99-1/100
=>A<99/100<1
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A=(1/2^2-1)(1/3^2-1)(1/4^2-1)...(1/100^2-1)
CM: A<-1/2
a) CM: A2= \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{100}}>10\)
b) CM: A3= \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}< 1\)
cm 1/2+2/2^2+3/2^3+4/2^4+...+100/2^100<2
Đặt \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\Leftrightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(\Leftrightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=2-\frac{100}{2^{99}}+\frac{100}{2^{100}}< 2-\frac{100}{2^{100}}+\frac{100}{2^{100}}=2\)
\(\Rightarrow A< 2\Leftrightarrow\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}< 2\left(đpcm\right).\)
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CM
1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2 < 1
Ta có:
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< 1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< 1\)
Vậy.......
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