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)
Tính:
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\)
Tính A = \(\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+...+\frac{n+2}{n!+\left(n+1\right)!+\left(n+2\right)!}\)
Tính :
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{n+1}\right)vớin\in N\)
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)
\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
\(\frac{1.3.5...\left(2n+1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1}{2^n}\) (n thuộc N*)
Chứng minh 2 câu trên
Tính:
\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
Tìm số nguyên N biết:
\(a,\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(b,\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(c,\left(\frac{-3}{4}\right)^n=\frac{81}{256}\)
\(d,\left(2x+3\right)^2=\frac{9}{121}^n\)
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<\(\frac{4}{3}\)