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Giải:
Ta có:
\(\dfrac{A}{B}=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{4026}}{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4025}}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2046}\right)}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}{1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{4025}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{4026}}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)
\(\Rightarrow\dfrac{A}{B}=1+\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2046}}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)
Dễ thấy \(\dfrac{A}{B}>1\)
Mà \(\dfrac{2013}{2014}< 1\)
\(\Rightarrow\dfrac{A}{B}>1\dfrac{2013}{2014}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
Ta có:
*) \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014}\right)\)
\(\Rightarrow S=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007}\right)\)
\(\Rightarrow S=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Vậy \(\left(S-B\right)^{2016}=\left[\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\right)\right]^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0^{2016}\)
\(\Rightarrow\left(S-B\right)^{2016}=0\)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2013.2014}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}=1-\dfrac{1}{2014}=\dfrac{2013}{2014}\left(đpcm\right)\)
Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.
Xét:
`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`
`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`
`=>A+B>2`
Mà `1 2013/2014<2`
`=>A+B>1 2013/2014`