Các câu hỏi tương tự
\(P=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)Khi đó \(\frac{2014}{2016}\).P=
\(P=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)Khi đó \(\frac{2014}{2016}\)P=
Cho \(P=\left(1-\frac{1}{1+2}\right)+\left(1-\frac{1}{1+2+3}\right)...\left(\frac{1}{1+2+..+2014}\right)\). Khi đó \(\frac{2014}{2016}P=\)
Cho \(P=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)....\left(1-\frac{1}{1+2+...+2014}\right)\)
Khi đó \(\frac{2014}{2016}P=...\)
Cho \(P=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+2014}\right)\)
Khi đó \(\frac{2014}{2016}P\)
Tính nhanh :
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)
Giúp mik nha
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)\)
Tính A=\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+3+...+2014}\right)\)