A=1/2²+1/3²+...+1/9²

Ta có:1/2²>1/2.3,1/3²>1/3.4,...,1/9²>1/9.10

⇒A>1/2.3+1/3.4+...+1/9.10

A>1/2-1/3+1/3-1/4+...+1/9-1/10

A>1/2-1/10

A>2/5(đpcm)

Ta có: A = 1/4 + 1/9 + 1/16 + 1/25 +1/36 + 1/49 + 1/64 + 1/81

Vì 1/2²>1/2.3,1/3²>1/3.4,...,1/9²>1/9.10

=>A>1/2.3+1/3.4+...+1/9.10

=>A>1/2-1/3+1/3-1/4+...+1/9-1/10

=>A>1/2-1/10

=>A>2/5

Giải:

\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{1}{64}+\dfrac{1}{81}\) 

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2} +\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\) 

\(\dfrac{1}{5^2}=\dfrac{1}{5.5}>\dfrac{1}{5.6}\) 

\(\dfrac{1}{6^2}=\dfrac{1}{6.6}>\dfrac{1}{6.7}\) 

\(\dfrac{1}{7^2}=\dfrac{1}{7.7}>\dfrac{1}{7.8}\) 

\(\dfrac{1}{8^2}=\dfrac{1}{8.8}>\dfrac{1}{8.9}\) 

\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\) 

\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\) 

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\) 

\(\Rightarrow A>\dfrac{2}{5}\left(đpcm\right)\) 

Chúc bạn học tốt!

ko bít làm thì thôi đi dễ quá mà

số số hạng của dãy số:

(10000-1):3+1=3334

tổng của dãy số là:

(10000+1).3334:2=16671667

k nha

1²+2²+3²+...+99²

Đặt    \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)

\(A=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)=\frac{1}{4}+\frac{1}{4}\cdot B\)

Ta có     \(\frac{1}{2^2}< \frac{1}{1\cdot2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)

\(...\)

\(\frac{1}{50^2}< \frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{4}\cdot1=\frac{1}{2}\)

Giúp đi

\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times\left(1-\frac{1}{25}\right)\times\left(1-\frac{1}{36}\right)\)

\(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times\frac{36}{36}\)

\(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times\frac{4.6}{5.5}\times\frac{5.7}{6.6}\)

\(=\frac{1.2.3.4.5}{2.3.4.5.6}\times\frac{3.4.5.6.7}{2.3.4.5.6}\)

\(=\frac{1}{6}\times\frac{7}{2}\)

\(=\frac{7}{12}\)

(1-1/4)×(1-1/9)×(1-1/16)×(1-1/25)×(1-1/36)

=(4/4-1/4)×(9/9-1/9)×(16/16-1/16)×(25/25-1/25)×(36/36-1/36)

=3/4×8/9×15/16×24/25×35/36

=1×3×2×4×3×5×4×6×5×7/2×2×3×3×4×4×5×5×6×6

=(1×2×3×4×5)×(3×4×5×6×7)/(2×3×4×5×6)×(2×3×4×5×6)

=1/6×7/2

=7/12

Thank you

\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....+\frac{1}{10000}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+.....+\frac{1}{100.100}\)

\(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}<\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}=\frac{99}{100}<1\)

Vậy \(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+.....+\frac{1}{10000}<1\)

  zxzXZda

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