Chứng minh rằng
1/32+1/42+1/52+1/62...+1/1002<1/2
Những câu hỏi liên quan
chung minh rang :
1/3^2+1/4^2+1/5^2+1/6^2+...+1/100^2<1/2
chung minh rang 1/3^2+1/4^2+1/5^2+...+1/100^2<1/2
có: 1/3^2<1/2.3; 1/4^2<1/3.4:...: 1/100^2<1/99.100
Mà: 1/1.2+1/2.3+...+1/99.100=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100
=99/100
=> 1/3^2+1/4^2+...+1/100^2<99/100<1
=> đpcm
UNDERSTAND ???
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đặt A= biểu thức trên
tao có
A<1/2.3+1/3.4+...+1/99.100
A<1/2-1/3+1/3-1/4+...+1/99-1/100
A<1/2-1/100<1/2
SUY RA A<1/2(DPCM)
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P=1/52+1/62+...+1/1002
chung minh rang 1/6<P<1/4
Ta có
\(P< \frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{99.100}\)
\(\Rightarrow P< \frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow P< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(\Rightarrow P< \frac{1}{4}\left(1\right)\)
\(p>\frac{1}{5^2}+\frac{1}{6.7}+....+\frac{1}{100.101}\)
\(P>\frac{1}{5^2}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(P>\frac{1}{6}+\frac{1}{25}-\frac{1}{101}\)
Ta thấy
\(\frac{1}{25}>\frac{1}{101}\Rightarrow\frac{1}{25}-\frac{1}{101}>0\)
Đặt \(M=\frac{1}{25}-\frac{1}{101}\)
\(\Rightarrow P>\frac{1}{6}+M>\frac{1}{6}\)
\(\Rightarrow P>\frac{1}{6}\left(2\right)\)
Tự (1) và (2)
\(\Rightarrow\frac{1}{6}< p< \frac{1}{4}\)
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chung minh rang : 1 / 2 ^ 2 + 1 / 3 ^ 2 + 1 / 4 ^ 2 + . . . + 1 / 100 ^ 2 < 99 / 100
Hình như sai đề thì phải chứ mk làm ko đc !!!
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A < 1/(1.2) + 1/(2.3) + 1/(3.4) + ...+ 1/(99.100)
<=> A< 1- 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + .. + 1/99 - 1/100
<=> A < 1 - 1/100 < 1 (đpcm)
So với thì đây
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Chung minh rang
S=1/5^2+1/6^2+1/7^2+....+1/100^2 nho hon 1/2
Ta có:
\(\frac{1}{5^2}<\frac{1}{4.5}\)
\(\frac{1}{6^2}<\frac{1}{5.6}\)
\(...\)
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{4}-\frac{1}{100}<\frac{1}{4}<\frac{1}{2}\)
Vậy \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}<\frac{1}{2}\)
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\(s=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
\(S=\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+...+\frac{1}{100.100}<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(S<\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow S<\frac{1}{5}-\frac{1}{101}\)
Vì \(\frac{1}{5}<\frac{1}{2}\)nên \(\frac{1}{5}-\frac{1}{101}<\frac{1}{2}\)
hay \(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{5}-\frac{1}{101}<\frac{1}{2}\)
Vậy \(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{2}\) (đpcm)
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bai 5 : tinh
a) tim x , biet (x+1) +(x+2 ) + ...+(x+100)=5750
b) chung minh rang B = 1/2^2 + 1/3^2 + 1/4^2 + ...+1/2021^2 < 1
giup mik luon voi
\(∘backwin\)
\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)
\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)
\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)
\(100 x + 5050 = 5750\)
\( 100 x = 5750 − 5050\)
\(100 x = 700\)
\(x = 700 : 100\)
\(x = 7\)
\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)
\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(B<1-\)\(\dfrac{1}{2021}\)
\(B<\)\(\dfrac{2020}{2021}\)
\(\dfrac{2020}{2021}< 1\)
\(B<1\)
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a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7
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b) Ta thấy: 1/2^2 < 1/2.3
1/3^2 < 1/3.4
...
1/2021^2 < 1/2021.2022
--> B=1/2^2 + 1/3^2 + 1/4^2 + ...+ 1/2021^2 < 1/2.3 + 1/3.4 + ... +1/2021.2022 (1)
Ta có: 1/2.3 + 1/3.4 + ... +1/2021.2022
=1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2021 - 1/2022
=1/2 - 1/2022 < 1 (2)
Từ (1) và (2) --> B<1 (đpcm)
<
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Xem thêm câu trả lời
Chung minh rang 1/22+1/42+1/62+...+1/1002<1/2
chung minh rang 1\2 mu 2+1\3 mu 2+1\4 mu2+...+1\100 mu 2 < 1
ai lam day du dau tien minh se k cho nha
minh can gap lam
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\).
chung minh rang 1-1/2^2-1/3^3-1/4^2-....-1/100^2>1/100
please,who can help me?
P=1/52+1/62+....+1/1002
chung minh rang:1/6<P<14
\(P=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
- Có: \(P>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
=> \(P>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
=> \(P>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)
=> \(P>\frac{1}{6}\)(1)
- Có: \(P< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
=> \(P< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
=> \(P< \frac{1}{4}-\frac{1}{100}< 14\)(2)
Từ (1) và (2)
=> \(\frac{1}{6}< P< 14\)(Nếu đề là 1/6 < P < 1/4 thì thay số 14 bằng 1/4 vẫn đúng nhé)
=> Đpcm
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