Các câu hỏi tương tự
CMR \(\frac{1.3.5.7............\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)............2n}\)=\(\frac{1}{2^n}\)
CMR : \(\frac{1.3.5.7..............\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...............2n}\) =\(\frac{1}{^{2^n}}\)
CMR \(\forall n\in\)N* ta có
\(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\)
1)CMR:
a) \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b) \(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)( n thuộc N* )
CMR
\(1\times3+2\times4+3\times5+\left(n-1\right)\left(n+1\right)=\frac{\left(n-1\right)n\left(2n+1\right)}{6}\)
Chứng minh rằng:
\(\frac{1.3.5.7.9.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{1}{2^n}\)
\(A=\frac{\left(n+1\right)\left(n+2\right)...2n}{2^n}\)
CMR: A là số tự nhiên với mọi n thuộc N.
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*
C/m rằng B= \(\frac{1.3.5............\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).......2n}=\frac{1}{2^2}\)