\(C=\left[1+\frac{1}{1\cdot3}\right]\left[1+\frac{1}{2\cdot4}\right]...\left[1+\frac{1}{2014\cdot2016}\right]\)

\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot...\cdot\frac{4060225}{4060224}\)

\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{2015\cdot2015}{2014\cdot2016}\)

\(=\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot2015\cdot2015}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot2014\cdot2016}\)

Để ý kĩ thì các thừa số dưới mẫu so với trên tử giống nhau chỉ khác 2016 nên C bằng:

C = 2*2*3*3*4*4*...*2015*2015/1*2*3*3*4*4*5*5*...*2015*2015*2016 = 1/2016

Ta có : (a-1)(a+1)=a²+a-a-1=a²-1

      \(\Rightarrow\)(a-1)(a+1)+1=a²

Từ đó ta có :

\(C=\frac{2^2}{1.3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{2015^2}{2014\cdot2016}\)

\(\Rightarrow\)\(C=\left(\frac{2\cdot3\cdot4\cdot...\cdot2015}{1\cdot2\cdot3\cdot...\cdot2014}\right)\cdot\left(\frac{2\cdot3\cdot4\cdot...2015}{3\cdot4\cdot5\cdot...\cdot2016}\right)\)

\(\Rightarrow\)\(C=\frac{2015}{1}\cdot\frac{1}{2016}\)

\(\Rightarrow\)\(C=\frac{2015}{2016}\)

\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)

\(\Rightarrow C=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2014.2016+1}{2014.2016}\)

\(\Rightarrow C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{2015^2}{2014.2016}\)

\(\Rightarrow C=\frac{\left(2.3.4.....2015\right)\left(2.3.4.....2015\right)}{\left(1.2.3.....2014\right)\left(3.4.5.....2016\right)}\)

\(\Rightarrow C=\frac{2015.2}{1.2016}=\frac{4030}{2016}=\frac{2015}{1008}\)

=1/2.(1+1/1.3).(1+1/2.4).(1+1/3.5)...(1+1/2014.2016)

=1/2.(1+1/1-1/3).(1+1/3-1/5)...(1+1/2014-1/2016)

=1/2.1+(1/1-1/2016)

=1/2.2015/2016

=2015/4032

sai roi

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\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{2015^2}{2014.2016}=\frac{\left(2.3.4......2015\right)}{\left(1.2.3......2014\right)}.\frac{\left(2.3.4.....2015\right)}{\left(3.4.5......2016\right)}=\frac{2015}{1}.\frac{2}{2016}=\frac{2015}{1008}\)

\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)

\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)

\(A=\frac{2.3.4...2015}{1.2.3...2014}.\frac{2.3.4...2015}{3.4.5...2016}\)

\(A=2015.\frac{1}{1008}\)

\(A=\frac{2015}{1008}\)

Ta có :

\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}............\frac{2015^2}{2014.2016}\)= \(\frac{2.2}{1.3}.\frac{3.3}{2.4}...........\frac{2015.2015}{2014.2016}=\frac{2.2015}{2016}=\frac{2015}{1008}\)

k cho mình nha

Ta có công thức :

\(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)

Áp dụng vào bài toán ta được :

\(C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}..........\frac{2015^2}{2014.2016}\)

\(=\frac{\left(2.3.4....2015\right)\left(2.3.4....2015\right)}{\left(1.2.3...2014\right)\left(3.4.5.....2016\right)}\)

\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)

=1(1/1*3*(1/2*4)*...*(1+1/2014*2016)

=1/2(2+2/1*3)+(2+2/2*4)*...(2+2/2014*2016)

=1/2(2+1/1-1/3)...(2+1/2014-1/2016)

=1/2*(1/1-1/2016)

=3023/4032

2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)

2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)

2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)

2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)

2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)

2A = \(\frac{4032}{2017}\)

A = \(\frac{4032}{2017}\)\(:2\)

A = \(\frac{2016}{2017}\)