CMR 1/2+1/2^2+...+1/2^100<1
LT
Những câu hỏi liên quan
CMR 100 - ( 1 + 1/2 + 1/3 + ... + 1/100 ) = 1/2 + 2/3 + ... + 99/100.
CMR
1-1/2^2-1/3^2-1/4^2-....-1/100^2<1/100
cmr 100-(1+1/2+1/3+...+1/100)=1/2+2/3+...+99/100
Ta có :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)=1.100-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+.......+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+.........+\frac{99}{100}\)
Vậy \(100-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+.....+\frac{99}{100}\left(ĐPCM\right)\)
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cmr
100-(1+1/2+1/3+...+1/100)=1/2+2/3+3/4+....+99/100
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
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CMR:(1+1/2+1/3+1/4+...+1/100)=1/2=2/3+3/4+...+99/100
CMR: A=1/2^2+1/3^2+1/4^2+...+1/100^2<1
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\Rightarrow A< 1\)
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A<11⋅2+12⋅3+13⋅4+...+199⋅100�<11⋅2+12⋅3+13⋅4+...+199⋅100
A<1−1100�<1−1100
99100<1⇒A<1
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CMR:
a)1/10^2 +1/11^2+1/12^2+...+1/100^2 >3/4
b)1/2^2+1/3^2+1/4^2+...+1/100^2<99/100
c)1/2^2+1/3^2+1/4^2+...+1/100^2<3/4
cmr :1/2^2+1/4^2+...+1/100^2<1
1/2!+1/3!+1/4!+...+1/100!<1
cmr 1/(2^2)+1/(4^2)+...+1/(100^2) < 1/2