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28 tháng 2 2021

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`

22 tháng 5 2017

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 \(\dfrac{A}{B}>1\)

\(\dfrac{2013}{2014}< 1\)

\(\Rightarrow\dfrac{A}{B}>1\dfrac{2013}{2014}\)

16 tháng 7 2023

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.

 

29 tháng 3 2018

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

\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)

25 tháng 7 2023

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

25 tháng 7 2023

Bạn xem lại đề 2, phần mẫu của N

24 tháng 3 2017

tất nhên là bằng 00000000000000000000000000000000000000

26 tháng 11 2017

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2014^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2013.2014}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}=1-\dfrac{1}{2014}=\dfrac{2013}{2014}\left(đpcm\right)\)