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\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.4}.....\frac{2015^2}{2014.2016}\)
\(=\frac{\left(2.3.4....2015\right)\left(2.3.4...2015\right)}{\left(1.2.3....2014\right)\left(3.4.5....2016\right)}\)
\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)
\(C=\left[1+\frac{1}{1\cdot3}\right]\left[1+\frac{1}{2\cdot4}\right]...\left[1+\frac{1}{2014\cdot2016}\right]\)
\(=\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot...\cdot\frac{4060225}{4060224}\)
\(=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot3}{2\cdot4}\cdot\frac{4\cdot4}{3\cdot5}\cdot...\cdot\frac{2015\cdot2015}{2014\cdot2016}\)
\(=\frac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot2015\cdot2015}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot2014\cdot2016}\)
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 :
\(C=\frac{2^2}{1.3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot...\cdot\frac{2015^2}{2014\cdot2016}\)
\(\Rightarrow\)\(C=\left(\frac{2\cdot3\cdot4\cdot...\cdot2015}{1\cdot2\cdot3\cdot...\cdot2014}\right)\cdot\left(\frac{2\cdot3\cdot4\cdot...2015}{3\cdot4\cdot5\cdot...\cdot2016}\right)\)
\(\Rightarrow\)\(C=\frac{2015}{1}\cdot\frac{1}{2016}\)
\(\Rightarrow\)\(C=\frac{2015}{2016}\)
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4.3.5...2014.2016}\)
\(=\frac{\left(2.3.4..2015\right)\left(2.3.4..2015\right)}{\left(1.2.3..2014\right)\left(3.4.5..2016\right)}\)
\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)
Vậy \(C=\frac{2015}{1008}\)
Có \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..........\)\(\left(1+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)....\left(\frac{2014.2016}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{2^2-1}{1.3}+\frac{1}{2.4}\right)\left(\frac{3^2-1}{2.4}+\frac{1}{2.4}\right)......\left(\frac{2015^2-1}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2015.2015}{2014.2016}\)
=\(\frac{2.2.3.3.....2015.2015}{1.3.2.4....2014.2015}\)
=\(\frac{\left(2.3...2015\right).\left(2.3.....2015\right)}{\left(1.2....2014\right).\left(3.4.....2016\right)}=\frac{2015.2}{2016}=\frac{4030}{2016}\)
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)+...+\left(1+\frac{1}{2014\cdot2016}\right)=\frac{x}{1008}\)
\(\Rightarrow\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot...\cdot\frac{4060225}{4060224}=\frac{x}{1008}\)
\(\Rightarrow\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2015\cdot2015\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2014\cdot2016\right)}=\frac{x}{1008}\)
\(\Rightarrow\frac{\left(2\cdot3\cdot4\cdot...\cdot2015\right)\left(2\cdot3\cdot4\cdot...\cdot2015\right)}{\left(1\cdot2\cdot3\cdot...\cdot2014\right)\left(3\cdot4\cdot5\cdot...\cdot2016\right)}=\frac{x}{1008}\)
\(\Rightarrow\frac{2015\cdot2}{1\cdot2016}=\frac{x}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\)
\(\Rightarrow x=2015\)
\(=\frac{4}{3}.\frac{9}{8}...\frac{4060225}{4060224}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3...2015.2015}{1.3.2.4...2014.2016}\)
\(=\frac{2.3...2015}{1.2...2014}.\frac{2.3...2015}{3.4...2016}\)
\(=2015.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}\)
\(=\frac{2015}{1008}\)
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