a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

\(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\)

\(2^2.A=1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{98}}\)

\(2^2.A-A=\left(1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+\frac{1}{2^8}+...+\frac{1}{2^{100}}\right)\)

\(4.A-A=1-\frac{1}{2^{100}}< 1\)

\(3A< 1\)

\(\Rightarrow A< \frac{1}{3}\left(đpcm\right)\)

Gọi dãy trên là A, Ta có: 

1/5²+1/6²+1/7²+...+1/100² < 1/4.5+1/5.6+1/6.7+...+1/99.100

<=> 1/5²+1/6²+1/7²+...+1/100² < 1/4 - 1/100

<=> 1/5²+1/6²+1/7²+...+1/100² < 6/25

Mà 6/25 < 1/4 => A < 1/4

6/25 > 1/6 => A > 1/6

V ậ y: 1/6 < A < 1/4

Ta thấy: k² > (k - 1)(k + 1)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right).\dfrac{1}{2}\)

\(=\left(1-\dfrac{1}{101}\right).\dfrac{1}{2}\)

\(=\dfrac{100}{101}.\dfrac{1}{2}< 1.\dfrac{1}{2}=\dfrac{1}{2}\)

ta có :\(\frac{1}{5^2}<\frac{1}{4.5}\)

 \(\frac{1}{6^2}<\frac{1}{5.6}\)

\(\frac{1}{7^2}<\frac{1}{6.7}\)

.....

\(\frac{1}{100^2}<\frac{1}{99.100}\)

\(\Rightarrow A<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

                \(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\)     (1)

Ta có : \(\frac{1}{5.6}<\frac{1}{5^2}\)'

\(\frac{1}{6.7}<\frac{1}{6^2}\)

....\(\frac{1}{100.101}<\frac{1}{100^2}\)

\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\) <A 

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{101}\) <A

\(\frac{1}{5}-\frac{1}{101}\) <A

mà \(\frac{96}{5.101}=\frac{96}{505}>\frac{96}{576}\)

hay \(A>\frac{1}{6}\)                                     (2)
từ (1); và (2) suy ra \(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+..+\frac{1}{100^2}<\frac{1}{4}\) (đpcm)

đây là cách dễ hiểu nhất nhé

bài này dễ lắm 8h30'  mình giải cho đang bận

1/4^2<1/3*4

1/5^2<1/4*5

...

1/100^2<1/99*100

=>A<1/3-1/4+1/4-1/5+...+1/99-1/100

=>A<1/3-1/100<1/3

\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

\(=\dfrac{1}{2^2}\cdot\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1+1=2\)

\(\Rightarrow\dfrac{1}{2^2}.\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)

\(\Rightarrow dpcm\)

\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{99}{200}< \dfrac{1}{2}\)

lúc mấy người nghĩ ra thì tui làm xong rồi bây h mách mk bài khác đi