Cmr: S= 1+1/2^2+1/3^2+1/4^2+...+1/100^2>3/2
Những câu hỏi liên quan
S=1+1/2^2+1/3^2+...+1/100^2
CMR S<2
Câu 2: CMR S<1/4 với S=1/4^2+1/6^2+...+1/(2n)^2
\(S=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
Mà \(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=2-\dfrac{1}{100}< 2\)
\(\Rightarrow\) \(S< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
Vậy \(S< 2\left(đpcm\right).\)
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Câu 1 :
Ta có :
\(S=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+..........+\dfrac{1}{100^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
........................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow S< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{99.100}\)
\(\Leftrightarrow S< 1+1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow S< 1+1-\dfrac{1}{100}\)
\(\Leftrightarrow S< 2+\dfrac{1}{100}< 2\)
\(\Leftrightarrow S< 2\rightarrowđpcm\)
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\(S=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
\(S< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(S< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(S< 2-\dfrac{1}{100}\)
\(S< 2\rightarrowđpcm\)
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CMR : S < 3 biết : 1 + 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
CMR
S= 1/2 mũ 2 +1/3 mũ 2 + 1/4 mũ 2 +...... + 1/100 mũ 2 <1
Cho `S=1/(5^2) + 2/(5^3) + 3/(5^4) + ... + 99/(5^100)` CMR `S<1/16`
Ta có :
`5S=5(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`5S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)`
`=>5S-S=1/5+2/(5^2)+3/(5^3)+...+99/(5^100)-(1/(5^2)+2/(5^3)+3/(5^4)+...+99/(5^100))`
`4S=1/5+1/(5^2)+1/(5^3)+1/(5^4)+...+1/(5^99) -99/(5^100)`
`20S=5(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`20S=1+1/5+1/(5^2)+....+1/(5^98)-99/(5^99)`
`=>20S-4S=(1+1/5+1/(5^2)+...+1/(5^98)-99/(5^99))-(1/5+1/(5^2)+1/(5^3)+...+1/(5^99)-99/(5^100))`
`=>16S=1-99/(5^99)-1/(5^99)-99/(5^100)`
Vì `-99/(5^99)-1/(5^99)-99/(5^100)<0=>1-99/(5^99)-1/(5^99)-99/(5^100)<1`
`=>S<1/16`
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S=1/5^2 - 2/5^3 + 3/5^4 -...+99/5^100-100/5^101
CMR S<1/3^6
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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
bai 1) tim x, y
x.y-x+2y=3
bai2 ) cmr
a)1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-4/3^4+....+99/3^99-100/3^100<3/16
c)1cho tổng gồm 2014 số hạng : s=1/4+2/4^2+3/4^3+.....2014/4^2014
a) tính giá trị của biểu thức D=3/1.3+3/3.5+3/5.7+...+3/49.51
b) CMR : S=1+1/2^2+1/3^2+1/4^2+...+1/100^2 <2
CMR:(1+1/2+1/3+1/4+...+1/100)=1/2=2/3+3/4+...+99/100