Các câu hỏi tương tự
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*
Chứng minh rằng :
a)\(\frac{1.3.5....9}{21.22.23....40}\)=\(\frac{1}{2^{20}}\)
b)\(\frac{1.3.5....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)=\(\frac{1}{2^2}\)
Chứng minh rằng:
a,\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b,\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)
Biết rằng n thuộc N*
Chứng minh rằng:
a) \(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b)\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1}{2^n}\)
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* )
cho \(A=\frac{7}{3}.\frac{37}{3^2}....\frac{6^{2n}+1}{3^{2n}}\)và \(B=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)...\left(1+\frac{1}{3^{2n}}\right)\)với n thuộc N
a) Chứng minh: 5A-2B là số tự nhiên
b) Chứng minh với mọi số tự nhiên n khác 0 thì 5A-2B chia hết cho 45
chứng minh rằng
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}\)=\(\frac{1}{2^n}\)
C/m rằng B= \(\frac{1.3.5............\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).......2n}=\frac{1}{2^2}\)
C/M rằng
a) \(\frac{1.3.5.....39}{21.22.23.......40}=\frac{1}{2^{10}}\)
b) \(\frac{1.3.5.....\left(2n-1\right)}{\left(n+1\right).\left(n+2\right)......2n}=\frac{1}{2^2}\)