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 \(P=\left(1-\frac{1}{1+2}\right)+\left(1-\frac{1}{1+2+3}\right)...\left(\frac{1}{1+2+..+2014}\right)\). 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 \(\frac{2014}{2016}P\)
P=(-2/1+2).(-2-3/1+2+3)...(-2-3-...-2014/1+2+...2014)
-P=(1.4/2.3)(2.5/3.4)...(2013.2016/2014.2015)
-P=(1.2.3...2014/2.3.4...2013)(4.5.6...2016/3.4.5...2015)
-P=(1/2014)(2016/3)
P=(-1/2014)(2016/3)
(2014/2016)P=-107/3021
Vay...
\(P=\left(1-\frac{1}{1+2}\right)\cdot\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\)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=
P=336/1007
2014/2016P=2014/2016.336/1007=1/3
\(A=\frac{\left(1-2\right).\left(1+2\right)}{2^2}.\frac{\left(1-3\right).\left(1+3\right)}{3^2}.......\frac{\left(1-2013\right).\left(1+2013\right)}{2013^2}.\frac{\left(1-2014\right).\left(1+2014\right)}{2014^2}\)
Tính A = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2014}\right)\left(1-\frac{1}{2015}\right)\left(1-\frac{1}{2016}\right)\)
Ta có :
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}\)
\(A=\frac{2.3.4.....2015}{2.3.4.....2015}.\frac{1}{2016}\)
\(A=\frac{1}{2016}\)
Vậy \(A=\frac{1}{2016}\)
Chúc bạn học tốt ~
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}\)
\(\Rightarrow A=\frac{1.2.3..2015}{2.3.4..2016}\)
\(\Rightarrow A=\frac{1}{2016}\)
k đúng cho mk nha
\(choP=\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2014}\right)\) thì \(\frac{2014}{2016}P=\)
tính tích:
\(\left(1-\frac{1}{2014}\right).\left(1-\frac{2}{2014}\right).\left(1-\frac{3}{2014}\right)...\left(1-\frac{2015}{2014}\right)\)
NHẤT ĐỊNH SẼ CÓ PHÂN SỐ \(1-\frac{2014}{2014}=0\)
NÊN tích dãy số đó là 0
tk nha
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
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{101}{102}=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}=\frac{C}{D}\)
Ta có: \(D=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)(có 2015 số hạng)
\(D=\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)-2015\)
\(D=2016+\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}-2015\)
\(D=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+1=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(D=2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=2016C\)
Vậy \(B=\frac{C}{D}=\frac{C}{2016C}=\frac{1}{2016}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{101}{102}=\frac{1\cdot2\cdot3\cdot....\cdot101}{2\cdot3\cdot4\cdot....\cdot102}\)
\(A=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}=\frac{1}{2016}\)