Tính:
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\)
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<\(\frac{4}{3}\)
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)
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)
Tinhs
\(\frac{\left(\frac{-1}{2}\right)^{2n}}{\left(\frac{-1}{2}\right)^n}\)
Chứng minh rằng:
\(a.A=\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+...+\frac{2n+1}{n^2\left(n+1\right)^2}< 1\)
\(b.B=\frac{1}{2}\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+...+\frac{1}{9240}\right)>\frac{57}{461}\)
Tính:
\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
Tính A = \(\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{n+2}{n!+\left(n+1\right)!+\left(n+2\right)!}\)
