Các câu hỏi tương tự
CMR
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng minh rằng :
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Chứng tỏ rằng:
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
Chứng minh rằng:
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}.\)
giải cụ thể ra giùm.
Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
Tính \(\frac{B}{A}\)biết
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}+...+\frac{1}{2008\cdot2009\cdot2010}\)
\(0.4\left(3\right)+0.6\left(2\right)\cdot2\frac{1}{2}\cdot\left[\left(\frac{1}{2+\frac{1}{3}}\right):0.5\left(8\right)\right]:\frac{50}{53}\)
Tính hợp lý
a)\(\left(-\frac{1}{3}\right)^2\cdot\left(5\frac{1}{2}\right)^0+\left(-\frac{1}{3}\right)^5\cdot4\frac{1}{2}\)
b) \(\left(\frac{2^2\cdot2^3}{4^2\cdot16}\right)^{15};\left(\frac{2^6}{16^2}\right)^{10}\)
c)\(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}\)
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}\)