Chứng minh rằng: \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
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Chứng minh rằng : \(\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+........+\dfrac{1}{2020^2}< \dfrac{1}{4}\)
\(A=\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1010^2}\right)\)
1/2^2+1/3^2+...+1/2010^2<1/1*2+1/2*3+...+1/2009*2010=1-1/2010<1
=>A<1/4
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CMR : a, \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
b, \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)
\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(...\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\) (**)
Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)
\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm
b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)
\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)
\(3VT=1-\dfrac{1}{64}< 1\)
\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)
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24 tháng 4 2023 lúc 8:59
1. chứng minh dfrac{1}{4}+ dfrac{1}{16}+ dfrac{1}{36}+dfrac{1}{64}+dfrac{1}{100}+dfrac{1}{144}+dfrac{1}{196}≤dfrac{1}{2}
2.tìm số nguyên n sao cho (3n +24) ⋮ (n-4)
cứu t zới!!!huhu
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1. chứng minh \(\dfrac{1}{4}\)+ \(\dfrac{1}{16}\)+ \(\dfrac{1}{36}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{100}\)+\(\dfrac{1}{144}\)+\(\dfrac{1}{196}\)≤\(\dfrac{1}{2}\)
2.tìm số nguyên n sao cho (3n +24) \(⋮\) (n-4)
cứu t zới!!!huhu
Giải thích các bước giải:
Đặt A= 1/4+1/16+1/36+1/64+1/100+1/144+1/196
= 1/2^2+ 1/4^2+ 1/6^2+….+ 1/16^2
= 1/2^2.( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)
Ta có 1+ 1/2^2+ 1/3^2+…+ 1/8^2< 1+ 1/1.2+ 1/2.3+….7.8= 1+ 1-1/2+ 1/2- 1/3+….+ 1/7- 1/8
= 2- 1/8< 2
Nên ( 1+ 1/2^2+ 1/3^2+…+ 1/8^2)< 2
=> A< 1/2^2 nhân 2= 1/2
Vậy A< 1/2
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Chứng minh rằng
S = \(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{36}\)+.........+\(\dfrac{1}{10000}\)<\(\dfrac{1}{2}\)
S = \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+......+\dfrac{1}{10000}\)
\(\Rightarrow S=\dfrac{1}{4.1}+\dfrac{1}{4.4}+\dfrac{1}{4.9}+.....+\dfrac{1}{4.2500}\)
\(\Rightarrow S=\dfrac{1}{4.\left(1+\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}\right)}< \dfrac{1}{2}\)
\(\RightarrowĐPCM\)
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Chứng tỏ rằng: \(\dfrac{1}{2}< \dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{9801}+\dfrac{1}{10000}< \dfrac{37}{50}\)
Ta có : \(\dfrac{1}{4}\)= \(\dfrac{1}{2.2}\)> \(\dfrac{1}{2.3}\)
\(\dfrac{1}{9}\)= \(\dfrac{1}{3.3}\)> \(\dfrac{1}{3.4}\)
\(\dfrac{1}{16}\)=\(\dfrac{1}{4.4}\)> \(\dfrac{1}{4.5}\)
.......
\(\dfrac{1}{9801}\)= \(\dfrac{1}{99.99}\)> \(\dfrac{1}{99.100}\)
\(\dfrac{1}{10000}\)= \(\dfrac{1}{100.100}\)> \(\dfrac{1}{100.101}\)
\(\Rightarrow\) \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+ ..... + \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\)> \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{99.100}\)+\(\dfrac{1}{100.101}\)
= \(\dfrac{3-2}{2.3}\)+ \(\dfrac{4-3}{3.4}\)+ \(\dfrac{5-4}{4.5}\) +...+ \(\dfrac{100-99}{99.100}\)+ \(\dfrac{101-100}{100.101}\)
= \(\dfrac{3}{2.3}\)- \(\dfrac{2}{2.3}\) + \(\dfrac{4}{3.4}\)-\(\dfrac{3}{3.4}\)+ \(\dfrac{5}{4.5}\)-\(\dfrac{4}{4.5}\)+...+ \(\dfrac{100}{99.100}\)- \(\dfrac{99}{99.100}\)+ \(\dfrac{101}{100.101}\)-\(\dfrac{100}{100.101}\)
= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+....+ \(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
= \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\) ; Mà \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\)= \(\dfrac{99}{202}\)< \(\dfrac{1}{2}\)
\(\Rightarrow\) \(\dfrac{1}{2}\)< \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...+ \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\) (1)
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Chứng mình rằng :
a) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \)\(\dfrac{1}{2}\)
b)\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\)
Giup mk nha ! Đang cần gấp lắm rùi !
Chứng minh:
a. \(A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
b.\(B=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}< \dfrac{3}{16}\)
c. \(C=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
18 tháng 3 2023 lúc 19:38
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
11 tháng 1 2022 lúc 15:04
Chứng tỏ rằng:
\(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{62}+\dfrac{1}{63}+\dfrac{1}{64}>4\)