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 !
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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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Bài 1: Chứng tỏ rằng :
dfrac{11}{15} dfrac{1}{21}+dfrac{1}{22}+......+dfrac{1}{60} dfrac{3}{2}
Bài 2: Chứng tỏ rằng:
dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+......+dfrac{1}{n^2} 1
dfrac{1}{4}+dfrac{1}{16}+dfrac{1}{36}+dfrac{1}{100}+dfrac{1}{144}+dfrac{1}{196} dfrac{1}{2}
dfrac{1}{5}+dfrac{1}{13}+dfrac{1}{25}+dfrac{1}{41}+dfrac{1}{61}+dfrac{1}{85}+dfrac{1}{113} dfrac{1}{2}
Đọc tiếp
Bài 1: Chứng tỏ rằng :
\(\dfrac{11}{15}< \dfrac{1}{21}+\dfrac{1}{22}+......+\dfrac{1}{60}< \dfrac{3}{2}\)
Bài 2: Chứng tỏ rằng:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......+\dfrac{1}{n^2}< 1\)
\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}< \dfrac{1}{2}\)
bài 2
| a;đặt biểu thức là S | |
| S < 1/1.2 + 1/2.3 + .......1/(n-1)n | |
| = 1- 1/2 +1 /2 -1/3+........ + 1/n-1 - 1/n | |
|
= 1 -1/n <1 |
|
| vậy S < 1 | |
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BT1: CMR:
a) dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{n^2} 1
b) dfrac{1}{4}+dfrac{1}{16}+dfrac{1}{36}+dfrac{1}{64}+dfrac{1}{100}+dfrac{1}{144}+dfrac{1}{196} dfrac{1}{2}
c) dfrac{1}{3}+dfrac{1}{30}+dfrac{1}{32}+dfrac{1}{35}+dfrac{1}{45}+dfrac{1}{47}+dfrac{1}{50} dfrac{1}{2}
d) dfrac{1}{2}-dfrac{1}{4}+dfrac{1}{8}-dfrac{1}{16}+dfrac{1}{32}-dfrac{1}{64} dfrac{1}{3}
e) 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}
f) d...
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BT1: CMR:
a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)
b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)
c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)
d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
BT2: Tính tổng
a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)
BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
CMR: 1 < S < 2
bài này có trong sách Nâng cao và Phát triển bạn nhé
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Chứng minh
a, A 1 + dfrac{1}{4} + dfrac{1}{16} + dfrac{1}{36} + dfrac{1}{64} + dfrac{1}{100} + dfrac{1}{144} + dfrac{1}{196} 1
b, B 1 + dfrac{1}{4} + dfrac{1}{9} + dfrac{1}{16} + .... + dfrac{1}{81} + dfrac{1}{100} 1dfrac{65}{132}
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Chứng minh
a, A = 1 + \(\dfrac{1}{4}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{36}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{100}\) + \(\dfrac{1}{144}\) + \(\dfrac{1}{196}\) < 1
b, B = 1 + \(\dfrac{1}{4}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{16}\) + .... + \(\dfrac{1}{81}\) + \(\dfrac{1}{100}\) > \(1\dfrac{65}{132}\)
Cho A = 1/2 .3/4.5/6.....199/200.Chứng tỏ rằng B mũ 2 <1/201.Bạn có làm dược ko ?
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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: \(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{36}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{100}\)+\(\dfrac{1}{144}\)+\(\dfrac{1}{196}\)< \(\dfrac{1}{2}\)
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{196}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{13^2}\)
Đặt \(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{13^2}\)\(<\)\(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\left(1\right)\)
Mà \(B=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{12\cdot13}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{12}-\dfrac{1}{13}\)
\(=\dfrac{1}{2}-\dfrac{1}{13}< \dfrac{1}{2}\left(2\right)\). Từ \((1)\) và \((2)\) ta có:
\(A< B< \dfrac{1}{2}\Rightarrow A< \dfrac{1}{2}\) (Điều phải chứng minh)
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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 , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Chứng minh rằng: \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)
Ta có:
\(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)
\(\Rightarrow A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)\)
\(\Rightarrow A< \dfrac{1}{4}.\dfrac{99}{50}\)
\(\Rightarrow A< \dfrac{99}{200}< \dfrac{1}{2}\)
Vậy \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\) (Đpcm)
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\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(2-\dfrac{1}{50}\right)< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
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