cho \(A=\frac{7}{3}.\frac{37}{3^2}....\frac{6^{2n}+1}{3^{2n}}\)và \(B=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)...\left(1+\frac{1}{3^{2n}}\right)\)với n thuộc N

a) Chứng minh: 5A-2B là số tự nhiên

b) Chứng minh với mọi số tự nhiên n khác 0 thì 5A-2B chia hết cho 45

a/Chứng minh rằng  \(\frac{2}{\left(2n+1\right)\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

b/Áp dụng chứng minh

\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{4003\left(\sqrt{2001}+\sqrt{2002}\right)}<\frac{2001}{2003}\)

Chứng minh rằng:

a)\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)

b)\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)với n thuộc N*

Cho n là số nguyên dương lớn hơn 1. Chứng minh rằng:

\(\frac{1}{n!}< \left(2-\frac{1}{n}\right)\left(2-\frac{3}{n}\right)...\left(2-\frac{2n-1}{n}\right)\)

Chứng minh rằng với \(n\in N\)* thì:

a, \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

b, \(1^3+2^3+3^3+...+n^3=\left(\frac{n\left(n+1\right)}{2}\right)^2\)

c, \(n+2\left(n-1\right)+3\left(n-2\right)+...+n=\frac{n\left(n+1\right)\left(n+2\right)}{6}\)

Chứng minh: \(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\)\(< \frac{1}{2}\)

a, chứng minh:

\(n^4+\frac{1}{4}=\left[\left(n-1\right)n+\frac{1}{2}\right].\left[\left(n+1\right)n+\frac{1}{2}\right]\)

b, Áp dụng câu a) thu gọn:

\(\frac{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right)...\left(13^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right)...\left(14^4+\frac{1}{4}\right)}\)

Chứng minh rằng:

\(\frac{1.3.5.7.9.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{1}{2^n}\)