K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

1/4^2+1/6^2+...+1/(2n)^2<1/4

=>1/2^2+1/3^2+...+1/n^2<1

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n-1\right)}=\dfrac{1}{2}-\dfrac{1}{n-1}< 1\)

=>ĐPCM

Đặt P= 1/4^2+1/6^2+1/8^2+...1/2n^2

= > P= 1/2.(2/2.4+2/4.6+2/6.8+...+ 2/(2n-2).2n)

=> P= 1/2.(1/2-1/2n)

=> P= 1/2.1/2-1/2.1/2n

=> P = (1/4 -1/2.1/2n)(1/4

Vậy P<1/4 ( đcpcm)

\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)

=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)

=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)

a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)

b: undefined

AH
Akai Haruma
Giáo viên
25 tháng 12 2018

Lời giải:

Ta có:

\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{(2n)^2}< \frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}(*)\)

Mà:

\(\frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n-1)(2n+1)}\)

\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+1}\right)\)

\(< \frac{1}{6}< \frac{1}{4}(**)\)

Từ \((*);(**)\Rightarrow N< \frac{1}{4}\) (đpcm)

10 tháng 6 2021

a) \(\dfrac{32}{\left(-2\right)^n}=4\)

\(\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3\)

=> n = 3

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=4=2^2\)

=> n = 2

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)

=> 2n - 1 = 3

=> 2n = 4

=> n = 2

10 tháng 6 2021

\(\left(-2\right)^3=-8\) bạn ạ chứ không phải là bằng 8 nên n = 3 là không đúng rồi 

18 tháng 5 2017

Ta có:\(\dfrac{1}{4^2}< \dfrac{1}{2.4}\)

\(\dfrac{1}{6^2}< \dfrac{1}{4.6}\)

\(\dfrac{1}{8^2}< \dfrac{1}{6.8}\)

...

\(\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right).2n}\)

=>\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2n-2\right)2n}=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\left(đpcm\right)\)

18 tháng 5 2017

Đặt A = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}\)

\(A=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\right)\)

Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)

Ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\) ( vì 1 > 0 ; 0 < 1.2 < 22 )

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\) ( vì 1 > 0 ; 0 < 2.3 < 32 )

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\) ( vì 1 > 0 ; 0 < 3.4 < 42 )

...

\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\) ( vì 1 > 0 ; 0 < ( n - 1 ) n < n2 )

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow B< 1-\dfrac{1}{n}< 1\Rightarrow A< 1.\dfrac{1}{4}\Rightarrow A< \dfrac{1}{4}\)

Tính giá trị biểu thức : 1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\) 2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\) 3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\) 4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\) 5. Cho...
Đọc tiếp

Tính giá trị biểu thức :

1. \(A=\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}\)

2. \(B=\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

3. \(C=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4\cdot6}\cdot\dfrac{5^2}{4\cdot6}\)

4. \(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right)\cdot\left(\dfrac{2}{3}\cdot\dfrac{1}{4}\right)^2\)

5. Cho \(M=8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\) ; \(N=\left(10\dfrac{2}{9}+2\dfrac{3}{5}\right)-6\dfrac{2}{9}\). Tính \(P=M-N\)

6. \(E=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)

7. \(F=\dfrac{\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{256}+\dfrac{3}{64}}{1-\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

8. \(G=\text{[}\dfrac{\left(6-4\dfrac{1}{2}\right):0,03}{\left(3\dfrac{1}{20}-2,65\right)\cdot4+\dfrac{2}{5}}-\dfrac{\left(0,3-\dfrac{3}{20}\right)\cdot1\dfrac{1}{2}}{\left(1,88+2\dfrac{3}{25}\right)\cdot\dfrac{1}{80}}\text{]}:\dfrac{49}{60}\)

9. \(H=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{4\cdot5\cdot6}+...+\dfrac{1}{98\cdot99\cdot100}\)

10. \(I=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)

11. \(K=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{999}\right)\)

12. \(L=1\dfrac{1}{3}+1\dfrac{1}{8}+1\dfrac{1}{15}...\) (98 thừa số)

13. \(M=-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{3}}}}\)

14. \(N=\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}\)

15. \(P=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{5}-1\right)...\left(\dfrac{1}{2001}-1\right)\)

16. \(Q=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2005\cdot2006}\right):\left(\dfrac{1}{1004\cdot2006}+\dfrac{1}{1005\cdot2005}+...+\dfrac{1}{2006\cdot1004}\right)\)

2
27 tháng 11 2017

1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)

2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)

3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)

4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)

29 tháng 4 2022

hôi lì sít

10 tháng 3 2019

\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)

7 tháng 4 2023

Ta có :22A=1+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{n^2}\)

            22A-A=1-\(\dfrac{1}{\left(2n\right)^2}\)

            3A=\(\dfrac{\left(2n\right)^2-1}{\left(2n\right)^2}\) <\(\dfrac{n^2}{\left(2n\right)^2}\)=\(\dfrac{1}{2}\)

          3A<\(\dfrac{1}{2}\) suy ra A<\(\dfrac{1}{2}\)

6 tháng 4 2023

A   = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+\(\dfrac{1}{\left(2.n\right)^2}\)

A  =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\)\(\dfrac{1}{\left(2.3\right)^2}\) +....+\(\dfrac{1}{\left(2.n\right)^2}\)

A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^2.2^2}\) + \(\dfrac{1}{2^2.3^2}\)+......+ \(\dfrac{1}{2^2.n^2}\)

A = \(\dfrac{1}{2^2}\) \(\times\) ( 1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+.......+ \(\dfrac{1}{n^2}\))

22 \(\times\) A = 1 + \(\dfrac{1}{2^2}\)\(\dfrac{1}{3^2}\)+......+\(\dfrac{1}{n^2}\)

     4A =  1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +......+ \(\dfrac{1}{n^2}\)

     4A = 1 + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ...+\(\dfrac{1}{n.n}\)

       1   = 1

     \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

      \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)

     ...................

 \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right).n}\)

Cộng vế với vế ta có: 

4A = 1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+....+\(\dfrac{1}{n.n}\) <1+ \(\dfrac{1}{1.2}\)\(\dfrac{1}{2.3}\)+ ......+ \(\dfrac{1}{\left(n-1\right).n}\)

4A < 1+ \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+....+\(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\) = 2 - \(\dfrac{1}{n}\)

A < ( 2 - \(\dfrac{1}{n}\)): 4 

A < 2 : 4 - \(\dfrac{1}{n}\) : 4

A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\)

Vậy A < \(\dfrac{1}{2}\)