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11 tháng 1 2018

Sửa đề : \(S=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{144}\right)\)

\(\Rightarrow S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{143}{144}\) 

\(\Rightarrow S=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{11.13}{12.12}\) 

\(\Rightarrow S=\frac{1.2.3.....11}{2.3.4.....12}.\frac{3.4.5.....13}{2.3.4.....12}\)  

\(\Rightarrow S=\frac{1}{12}.\frac{13}{2}\) 

\(\Rightarrow S=\frac{13}{24}\)

9 tháng 8 2020

\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)

\(\Rightarrow P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)

Ta có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{12}\)

\(\Rightarrow P< \frac{2}{3}\left(đpcm\right)\)

9 tháng 8 2020

\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)

\(P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)

Có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow P< \frac{1}{4}=\frac{1}{2}-\frac{1}{12}\)

\(\Rightarrow P< \frac{2}{3}\)( đpcm )

`= 3/4 . 8/9 . 15/16. ... . 143/144`

`= (1.2.3...12)/(2.3.4....12) . (3.4.....12.13)/(2.3....11.12)`

`= 1/12 . 13/2 = 13/24`

=(1-1/2)(1+1/2)(1-1/3)(1+1/3)*...*(1-1/12)(1+1/12)

=1/2*2/3*...*11/12*3/2*4/3*...*13/12

=1/12*13/2=13/24

1: 8=2^3

2: 25=5^2

3: 4=2^2

4: 49=7^2

5: 81=9^2

6: 36=6^2

7: 100=10^2

8: 121=11^2

9: 144=12^2

10: 169=13^2

11: 27=3^3

12: 125=5^3

13: 1000=10^3

14: 32=2^5

15: 243=3^5

16: 343=7^3

17: 216=6^3

18: 64=4^3

19: 225=15^2

20: 128=2^7

29 tháng 7 2023

Giúp mình bài tiếp với ạ

31 tháng 1 2016

S=1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81=1-1/81=1/81

 

31 tháng 1 2016

80/81 là đúng

14 tháng 6 2016

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{120}{121}=\frac{3.8.15...120}{4.9.16...121}\)

    \(=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)...\left(11.11\right)}\)

    \(=\frac{\left(1.2.3...10\right).\left(3.4.5...12\right)}{\left(2.3.4...11\right).\left(2.3.4...11\right)}=\frac{1.12}{11.2}=\frac{6}{11}\)

14 tháng 6 2016

ta có :

A=\(\left(-\frac{3}{4}\right)\left(-\frac{8}{9}\right)\left(-\frac{15}{16}\right)...\left(-\frac{120}{121}\right)\)(có 10 số hạng)

  = \(\frac{3\cdot8\cdot15\cdot...\cdot120}{4\cdot9\cdot16\cdot...\cdot121}=\frac{\left(1.3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(10\cdot12\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot11^2}=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot..\cdot11\right)\left(2\cdot3\cdot4\cdot..\cdot11\right)}\)

=\(\frac{12}{11\cdot2}=\frac{12}{22}\)