Thu Thảo Vũ tick đúng cho mình nhé Thu Thảo Vũ

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

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

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

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

\(3A< 1\)

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

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

Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow2^2A=1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\)

\(\Rightarrow2^2A-A=\left(1+\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+...+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^4}+\dfrac{1}{2^6}+\dfrac{1}{2^8}+...+\dfrac{1}{2^{100}}\right)\)

\(\Rightarrow3A=1-\dfrac{1}{2^{100}}\)

\(\Rightarrow A=\dfrac{1-\dfrac{1}{2^{100}}}{3}< \dfrac{1}{3}\)(đpcm)

this sentence extremely easy

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