Cho C=1/1^2+1/2^2....1/n^2.Chứng tỏ 1<C<2
Những câu hỏi liên quan
1,Cho A=1/22+1/32+1/42+.............+1/1002
Chứng tỏ 1/2<A<3/4
2,Cho a,b,c \(\in\)N*,A=a/a+b+b/b+c+c/c+a
Chứng tỏ A không phải là số nguyên
Cho n thuộc N* chứng tỏ rằng : 1/2^2+1/3^2+...+1/n^2
1/2^2<1/1*2
1/3^2<1/2*3
...
1/n^2<1/(n-1)*n
=>1/2^2+1/3^2+...+1/n^2<1-1/2+1/2-1/3+...+1/n-1-1/n=1-1/n=(n-1)/n<1
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3) Cho C = 1/1 mũ 2 + 1/2 mũ 2 + 1/3 mũ 2 + … + 1/50 mũ 2. Chứng tỏ C < 173/100
4) Cho D = 3/4 + 8/9 + 15/16 +…+ 2499/2500. Chứng tỏ D > 48
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Cho F=1/1^2+1/2^2+1/3^2+1/4^2+...1/(n-1)+1/n^2,(n thuộc N*(.Chứng tỏ rằng :F<2
Cho tổng S = 1/2^2 + 1/3^2 +1/4^2 +... + 1/(n - 1)^2 +1/n^2. Chứng tỏ rằng S<1
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
Vậy S<1
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Ta có :
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-2\right)\left(n-1\right)}+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-2}-\frac{1}{n-1}+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow S< 1-\frac{1}{n}< 1\)
Vậy \(S=1\)
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cho n thuộc N*, chứng tỏ 1/1^2+1/2^2+1/3^2+...+1/n^2 không phải là một số tự nhiên
Đặt \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{n^2}\)
Có \(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{\left(n-1\right).n}\)
\(< -1.\left(\frac{1}{n}\right)< 1.\left(\frac{1}{n}\right)>0\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}< \frac{1}{1^2}+1< \orbr{\begin{cases}1+1\\2\end{cases}}\)
Vậy ta có điều phải chứng tỏ
Cho n thuộc N sao hãy chứng tỏ :
1/n(n+1)(n+2)=1/2[1/n(n+1)-1/(n+1)(n+2)]
Bài 1:
a, Cho A dfrac{1}{2^2}+dfrac{1}{4^2}+dfrac{1}{6^2}+....+dfrac{1}{100^2}
Chứng tỏ: A dfrac{1}{2}
b, Cho B dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+....+dfrac{1}{2^{20}}
Chứng tỏ B 1
c, Cho C dfrac{1}{5}+dfrac{1}{13}+dfrac{1}{14}+dfrac{1}{15}+dfrac{1}{61}+dfrac{1}{62}+dfrac{1}{63}
Chứng tỏ C dfrac{1}{2}
d, Cho D dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{20^2}
Chứng tỏ D 1
Đọc tiếp
Bài 1:
a, Cho A = \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}\)
Chứng tỏ: A <\(\dfrac{1}{2}\)
b, Cho B = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{20}}\)
Chứng tỏ B < 1
c, Cho C = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
Chứng tỏ C < \(\dfrac{1}{2}\)
d, Cho D = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}\)
Chứng tỏ D < 1
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
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A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
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!!!
1.
a, chứng tỏ
1/2^2+1/3^2+...+1/2017^2<1
b,1/4+1/16+1/36+1/64+1/100+1/144+...+1/10000<1/2
c,cho A=1/2^2+1/3^2...+1/9^2
chứng tỏ:2/5<a<8/9
d,chứng tỏ:A=1+1/2^2+...+1/100^2<1/3/4
e,chứng tỏ:1/2^2+1/3^2+...+1/100^2<1
a, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)Vậy...
b, Đặt A = \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
Thay B vào A ta được:
\(A< \frac{1}{4}\left(1+1\right)=\frac{1}{4}.2=\frac{1}{2}\)
Vậy....
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c, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};....;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(1)
Lại có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) suy ra \(\frac{2}{5}< A< \frac{8}{9}\)(đpcm)
d, chắc là đề sai
e, giống câu a
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