chung minh rang \(\frac{1\cdot3\cdot5\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\frac{1}{2^{20}}\)
\(\frac{1\cdot3\cdot5\cdot...\cdot39}{21\cdot22\cdot23\cdot...\cdot40}\)=\(\frac{1}{2^{20}}\)
\(\frac{1.3.5....39}{21.22.23.....40}=\frac{\left(1.3.5.....39\right).\left(2.4.6.....40\right)}{\left(21.22.....40\right).\left(2.4.6.....40\right)}\)
\(=\frac{1.2.3.4.....40}{21.22.....40.\left(1.2.3.....40\right)2^{20}}\)
\(=\frac{1}{2^{20}}\)
Chứng minh rằng
\(\frac{1\cdot3\cdot5\cdot\cdot\cdot\left(2n-1\right)}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)\cdot...\cdot2n}=\frac{1}{2^n}\)
Chứng minh:
\(\frac{1\cdot3\cdot5\cdot...\cdot39}{21\cdot22\cdot23\cdot...40}\)=\(\frac{1}{2^{20}}\)
Chứng minh rằng :
\(\frac{1\cdot3\cdot5\cdot...\cdot\left(2n-1\right)}{\left(n+1\right)\cdot\left(n+2\right)\cdot\left(n+3\right)\cdot...\cdot2n}=\frac{1}{2^n}\)
Tính
A=\(\left(1-\frac{1}{21}\right)\cdot\left(1-\frac{1}{28}\right)\cdot\left(1-\frac{1}{36}\right)\cdot....\cdot\left(1-\frac{1}{1326}\right)\)
B=\(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\frac{1}{99\cdot101}\right)\)
\(C=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot....\cdot\left(1-\frac{2}{99\cdot100}\right)\)
A = \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot.....\cdot\left(1+\frac{1}{2011\cdot2013}\right)\)
G=\(\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{50^2}{49.51}\)
H=\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{10}{7}\right)\)
Giúp mình vs
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
\(G=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
\(=\frac{\left(2.3.4.....50\right).\left(2.3.4.....50\right)}{\left(1.2.3.....49\right).\left(3.4.5.....51\right)}\)
\(=\frac{50.2}{51}=\frac{100}{51}\)
\(H=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....\left(1-\frac{7}{7}\right).....\left(1-\frac{10}{7}\right)\)
\(=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right).\left(1-\frac{3}{7}\right).....0.....\left(1-\frac{10}{7}\right)\)
\(=0\)
