\(A=\left(\frac{1}{1+2}\right).\left(\frac{1}{1+2+3}\right).....\left(\frac{1}{1+2+3+...+2014}\right)\)

\(A=\left(\frac{1}{\frac{2.\left(2+1\right)}{2}}\right).\left(\frac{1}{\frac{3.\left(3+1\right)}{2}}\right).....\left(\frac{1}{\frac{2014.\left(2014+1\right)}{2}}\right)\)

\(A=\frac{1}{\frac{2.3}{2}}.\frac{1}{\frac{3.4}{2}}.\frac{1}{\frac{4.5}{2}}.....\frac{1}{\frac{2014.2015}{2}}\)

\(A=\frac{2}{2.3}.\frac{2}{3.4}.\frac{2}{4.5}.....\frac{2}{2014.2015}\)

Đến đây thì không tính được nữa , có thể bạn chép nhầm dấu cộng thành dấu nhân rồi.

Nếu đổi dấu nhân thành dấu cộng, ta được:

\(A=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2014.2015}\)

\(A=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2014.2015}\right)\)

\(A=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{2014}-\frac{1}{2015}\right)\)

\(A=2.\left(\frac{1}{3}-\frac{1}{2015}\right)\)

\(A=2.\frac{2012}{6045}\)

\(A=\frac{4024}{6045}\)

\(A=\left(\frac{5}{1.2.3}+\frac{5.2}{2.3.4}+....+\frac{5.2014}{2014.2015.2016}\right)+\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{2014.2015.2016}\right)\)

\(A=\left(\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{2015.2016}\right)+\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)

\(A=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)+\frac{1}{2}-\frac{1}{2015}+\frac{1}{2016}\)

\(A=\frac{5}{2}-\frac{5}{2016}+\frac{1}{2}-\frac{1}{2015}+\frac{1}{2016}=3-\frac{1}{504}-\frac{1}{2015}\)

NHẤT ĐỊNH SẼ CÓ PHÂN SỐ \(1-\frac{2014}{2014}=0\)

NÊN tích dãy số đó là 0

tk nha

100 ngày

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

\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)( có 2013 thừa số ) 

\(A=\left(-\frac{3}{2^2}\right).\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right).....\left(-\frac{\text{4056196}}{2014^2}\right)\)

\(-A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{4056196}{2014^2}=\frac{1.3.2.4.3.5....2013.2015}{2.2.3.3.4.4.....2014.2014}\)

\(-A=\frac{\left(1.2.3...2013\right).\left(3.4.5.6...2015\right)}{\left(2.3.4.5....2014\right).\left(2.3.4.5...2014\right)}=\frac{1.2015}{2.2014}=\frac{2015}{4028}\)

\(A=-\frac{2015}{4028}\)

Vậy.....

-A=(\(1-\frac{1}{2^2}\)) . (\(1-\frac{1}{3^2}\))......(\(1-\frac{1}{2014^2}\))

-A= \(\frac{3}{4}\). \(\frac{8}{9}\). ...... \(\frac{4056195}{4056196}\)

-A= \(\frac{1.3.2.4.......2013.2015}{2.2.3.3.......2.14.2014}\)

-A= \(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\)

-A= \(\frac{2015}{2014.2}\)

-A=\(\frac{2015}{4028}\)

mình quên A=\(\frac{-2015}{4028}\)

Ta có:\(\left(x-1\right)\left(x+1\right)=x\left(x-1\right)+x-1^2=x^2-x+x-1=x^2-1\)

Áp dụng:\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)

                  \(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{2014^2-1}{2014\cdot2014}\)

                  \(=\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot...\cdot\frac{2013\cdot2015}{2014^2}\)

                  \(=\frac{1}{2}\cdot\frac{2015}{2014}=\frac{2015}{4028}\)