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

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

\(...\)

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

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

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

1/4 = 1/(2*2) < 1/(1*2) = 1/2 - 1/4 
tương tự ta có 
1/16 < 1/(2*4) = 1/4 - 1/8 
1/36 < 1/(4*6) = 1/8 - 1/12 
1/64 < 1/(6*8) = 1/12 - 1/16 
1/100 < 1/(8*10) = 1/16 - 1/20 
1/144 < 1/(10*12) = 1/20 - 1/24 
1/196 < 1/(12* 14) = 1/24 - 1/28 
cộng hết lại 
=> 1/4 + 1/16 + ......+ 1/100 + 1/144 + 1/196 < 1/2 - 1/28 < 1/2 => đpcm

ta có 
1/4 = 1/(2*2) < 1/(1*2) = 1/2 - 1/4 
tương tự ta có 
1/16 < 1/(2*4) = 1/4 - 1/8 
1/36 < 1/(4*6) = 1/8 - 1/12 
1/64 < 1/(6*8) = 1/12 - 1/16 
1/100 < 1/(8*10) = 1/16 - 1/20 
1/144 < 1/(10*12) = 1/20 - 1/24 
1/196 < 1/(12* 14) = 1/24 - 1/28 
cộng hết lại 
=> 1/4 + 1/16 + ......+ 1/100 + 1/144 + 1/196 < 1/2 - 1/28 < 1/2 => đpcm
Tick đúng nha bạn
 

1/4+1/16+1/36+...+1/196<1/4.2=1/2

trong nang cao va phat trien toan 6

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!

Ta có:

1/2 + 1/3 + 1/4 + ... + 1/15 + 1/16 = (1/2 + 1/3 + 1/4 + 1/5) + (1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11) + (1/12 + 1/13 + 1/14) + (1/15 + 1/16)

Vì 1/6 + 1/7 + 1/8 < 3x 1/6 = 1/2

   1/9 + 1/10 + 1/11 <3x1/9 = 1/3

   1/12 + 1/13 +1/14 < 3x1/12 = 1/4

   1/15 + 1/16 < 3 x 1/15 = 1/5

Nên A < 2 x (1/2 + 1/3 + 1/4 + 1/5) < 2 x (1/2 + 1/2 + 1/4 + 1/4) =3 (1)

Lập luận tương tự có:

A = ( 1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12) + (1/13 + 1/14 + 1/15 + 1/16) > (1/2 + 1/3 + 1/4) + 4 x 1/8 + 4 x 1/ 12 + 4 x 1/16

Hay A > 2 x (1/2 + 1/3 + 1/4) > 2 x (1/2 + 1/4 + 1/4) = 2 (2)

Từ (1) và (2) ta có 2 < A < 3. Vậy A không phải là số tự nhiên.

Làm piếng viết phân số nên bạn lm đỡ nhé!!!!!!!!!!!!!!

Ta có

\(A=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}\right)\)

Vì \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}< \frac{1}{6}.3=\frac{1}{2}\)

    \(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}< \frac{1}{9}.3=\frac{1}{3}\)

   \(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}< \frac{1}{12}.3=\frac{1}{4}\)

   \(\frac{1}{15}+\frac{1}{16}< \frac{1}{10}.2=\frac{1}{5}\)

=> \(S< 2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)< 2\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=3\)

=> S<3 (1) 

Lập luận tương tự ta có

\(S>2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)>2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=2\)

=> S>2 (2)

Từ (1) và (2) ta có 2 < A < 3. Vậy A không phải là số tự nhiên.