\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
\(\frac{1.3.5...\left(2n+1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1}{2^n}\) (n thuộc N*)
Chứng minh 2 câu trên
Chứng minh rằng:
\(a.A=\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...+\frac{2n+1}{n^2\left(n+1\right)^2}< 1\)
\(b.B=\frac{1}{2}\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}\right)>\frac{57}{461}\)
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, \(n+2\left(n-1\right)+3\left(n-2\right)+...+n=\frac{n\left(n+1\right)\left(n+2\right)}{6}\)
Tinhs
\(\frac{\left(\frac{-1}{2}\right)^{2n}}{\left(\frac{-1}{2}\right)^n}\)
Tính:
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\)
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)
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)
Bài 1:
a)Tìm x:
\(\frac{x+4}{2008}+\frac{x+3}{2009}=\frac{x+2}{2010}+\frac{x+1}{2011}\)
b) cho: \(M=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}\)
Bài 2:
a) Tìm x,y biết:
\(\frac{x+2y}{18}=\frac{1+4y}{24}=\frac{1+x+6y}{6x}\)
b) tìm ssos nguyên n để A mang giá trị nguyên và tính giá trị đó
\(A=\frac{9+3n}{n-4}\)
Tính B = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+4+...+16\right)\)
