Ta có:
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}=\frac{\left(1.3.5...2n-1\right).\left(2.4.6...2n\right)}{\left(2.4.6...2n\right)\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\)
\(=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{1.2.3...n\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n.2^n}\)
\(=\frac{1}{2^n}\)
\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
\(\frac{1.3.5...\left(2n+1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1}{2^n}\) (n thuộc N*)
Chứng minh 2 câu trên
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, \(n+2\left(n-1\right)+3\left(n-2\right)+...+n=\frac{n\left(n+1\right)\left(n+2\right)}{6}\)
Rút gọn biểu thức sau:
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\) (n nguyên dương)
Tinhs
\(\frac{\left(\frac{-1}{2}\right)^{2n}}{\left(\frac{-1}{2}\right)^n}\)
Cho M=\(\frac{1.3+2}{4}.\frac{3.5+2}{16}.\frac{15.17+2}{256}.\frac{255.257+2}{65536}.....\frac{\left(2^{2n}-1\right)\left(2^{2n}+1\right)+2}{2^{2n}}\)
(n thuộc N)
Chứng minh M<\(\frac{4}{3}\)
Tìm \(n\in N\) để:
a, \(\left(2n+1\right)⋮\left(6-n\right)\)
b, \(\left(3n+1\right)⋮\left(11-2n\right)\)
Chứng minh rằng với mọi số nguyên dương n thì:
\(5n=1^2+2^2+3^2+...+n^2=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)\)
(quy nạp)
cm biểu thức
\(n\left(2n-3\right)-2n\left(n+1\right)⋮5\) với mọi số nguyên n
Tính:
\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
