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\(C=\left[1+\frac{1}{1\cdot3}\right]\left[1+\frac{1}{2\cdot4}\right]...\left[1+\frac{1}{2014\cdot2016}\right]\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot...\cdot\frac{4060225}{4060224}\)
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{2015\cdot2015}{2014\cdot2016}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot2015\cdot2015}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot2014\cdot2016}\)
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 :
\(C=\frac{2^2}{1.3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{2015^2}{2014\cdot2016}\)
\(\Rightarrow\)\(C=\left(\frac{2\cdot3\cdot4\cdot...\cdot2015}{1\cdot2\cdot3\cdot...\cdot2014}\right)\cdot\left(\frac{2\cdot3\cdot4\cdot...2015}{3\cdot4\cdot5\cdot...\cdot2016}\right)\)
\(\Rightarrow\)\(C=\frac{2015}{1}\cdot\frac{1}{2016}\)
\(\Rightarrow\)\(C=\frac{2015}{2016}\)
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.4}.....\frac{2015^2}{2014.2016}\)
\(=\frac{\left(2.3.4....2015\right)\left(2.3.4...2015\right)}{\left(1.2.3....2014\right)\left(3.4.5....2016\right)}\)
\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)+...+\left(1+\frac{1}{2014\cdot2016}\right)=\frac{x}{1008}\)
\(\Rightarrow\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot...\cdot\frac{4060225}{4060224}=\frac{x}{1008}\)
\(\Rightarrow\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2015\cdot2015\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2014\cdot2016\right)}=\frac{x}{1008}\)
\(\Rightarrow\frac{\left(2\cdot3\cdot4\cdot...\cdot2015\right)\left(2\cdot3\cdot4\cdot...\cdot2015\right)}{\left(1\cdot2\cdot3\cdot...\cdot2014\right)\left(3\cdot4\cdot5\cdot...\cdot2016\right)}=\frac{x}{1008}\)
\(\Rightarrow\frac{2015\cdot2}{1\cdot2016}=\frac{x}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\)
\(\Rightarrow x=2015\)
https://dethihsg.com/de-thi-hoc-sinh-gioi-phong-gddt-hoang-hoa-2014-2015/
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\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)
\(A=\frac{2.3.4...2015}{1.2.3...2014}.\frac{2.3.4...2015}{3.4.5...2016}\)
\(A=2015.\frac{1}{1008}\)
\(A=\frac{2015}{1008}\)