What the heck! Bài lớp 6 khó thiệt! Phải chuẩn bị tinh thần thoi!

nhanh giúp mình

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{100}=\frac{99}{100}< 1< 2\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{100^2}< \frac{1}{99.100}\)

=> \(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

=> \(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(A< 1+1-\frac{1}{100}\)

=> \(A< 2-\frac{1}{100}< 2\)

Vậy \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 2\)(đpcm)

B1 : B-A = 1/2

B2 : 

CM được : A = (4^100-1)/3

=> A < 4^100/3 = B/3

Tk mk nha

Bài 1 :

A = 1 + 3 + 3² + 3³ + ....... + 3²⁰

\(\Rightarrow3A=3+3^2+3^3+3^4+......+3^{21}\)

\(\Rightarrow3A-A=\left(3+3^2+3^3+3^4+.....+3^{21}\right)-\left(1+3+3^2+3^3+......+3^{20}\right)\)

\(\Rightarrow2A=2+3^{21}\)

\(\Rightarrow A=\frac{2+3^{21}}{2}\)

\(\Rightarrow B-A=\left(2+3^{21}\right):2-3^{21}:2\)

\(\Rightarrow B-A=1+3^{21}:2-3^{21}:2\)

\(\Rightarrow B-A=1+\left(3^{21}:2-3^{21}:2\right)\)

\(\Rightarrow B-A=1+0\)

\(\Rightarrow B-A=1\)

Vậy \(B-A=1\)

Bài 2 : 

\(A=1+4+4^2+4^3+.....+4^{99}\)

\(\Rightarrow4A=4+4^2+4^3+4^4+.....+4^{100}\)

\(\Rightarrow4A-A=\left(4+4^2+4^3+4^4+.....+4^{100}\right)-\left(1+4+4^2+4^3+......+4^{99}\right)\)

\(\Rightarrow3A=3+4^{100}\)

\(\Rightarrow A=\frac{3+4^{100}}{3}\)

\(\Rightarrow\frac{B}{3}=\frac{4^{100}}{3}\)

Vì \(4^{100}=4^{100}\)nên \(3+4^{100}>4^{100}\)

Vậy \(A>\frac{B}{3}\left(ĐPCM\right)\)