Các câu hỏi tương tự
tính \(D=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)...\left(1-\frac{1}{\left(2n-1\right)^2}\right)\)n thuộc N, n>1
\(D=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{1}{\left(2n-1\right)^2}\right),\)với \(n\in N,n\ge1\)
CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
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*
1. Chứng minh rằng với n là stn khác 0 thì \(4^{2n+1}+3^{n+2}\)chia hết cho 13.
2.Tính:
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{n+1}\right)\)
Tính:
\(F=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{n^2}-1\right)\)
cho \(I=\frac{1.3+2}{4}.\frac{3.5+2}{16}.....\frac{\left(2^{2n}-1\right)\left(2^{2n}+1\right)+2}{2^{2n}}\)với n thuộc N. chứng minh \(I< \frac{4}{3}\)
Tìm n thuộc N, biết: \(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)...2n}\frac{1}{2^n}\)
Tính Q=\(\frac{1.3}{3.5}+\frac{2.4}{5.7}+\frac{3.5}{7.9}+.....+\frac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+......+\frac{1002.1004}{2005.2007}\)