(1+3+5+7+...+2019+2021)

A=1−3+5−7+......−2019+2021−2023

A=(1−3)+(5−7)+....+(2021−2023)A=(1−3)+(5−7)+....+(2021−2023)

A=−2+(−2)+....+(−2)(506)A=−2+(−2)+....+(−2)(506cặp)

a=−2.506A=−2.506

A=−1012A=−1012

(2+4+6+8+...+2020)

B=2+4+6+8+...+2018+2020
B = 2(1 + 2 + 3 + 4 + ... + 1009 + 1010)
B = 2 . (1011 . 1010 : 2)
B = 2 . 510555
B = 1 021 110

(1+3+5+7+......+2019+2021)-(2+4+6+8+.....+2020)

\(=\dfrac{\left(1+2021\right).\left[\left(2021-1\right):2+1\right]}{2}-\dfrac{\left(2+2020\right).\left[\left(2020-2\right):2+1\right]}{2}\)

\(=1011\)

Nhanh giúp mk nhé!

Cần gấp lắm!

số lượng số hạng của dãy số là 

    (  2021 - 2  ) : 1 + 1 = 2020 

tổng của dãy số là 

  ( 2021 + 2) x 2020 : 2 = 2043230

                                     vậy A = \(\frac{1}{2043230}\)

xong a rồi vậy b thì sao bạn

Ta có : 1² - 2² + 3² - 4² +  5² - 6² + .... + 2019² - 2020²

= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + .... + (2019 - 2020)(2020 + 2019)

= -3 - 7 - 11 - ...  - 4039

= - (3 + 7 + 11 + ... + 4039)

= - 1010.(4039 + 3) : 2 

= - 1010.2021

= -2041210

\(=\left(2^2-1\right)+\left(4^2-3^2\right)+\left(6^2-5^2\right)+...+\left(2020^2-2019^2\right)=\)

\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2020-2019\right)\left(2020+2019\right)=\)

\(=3+7+11+....+4039=\frac{1009\left(4039+3\right)}{2}=\)

1² - 2² + 3² - 4² + ... + 2019² - 2020²

= ( 1² - 2² ) + ( 3² - 4² ) + ... + ( 2019² - 2020² )

= ( 1 - 2 )( 1 + 2 ) + ( 3 - 4 )( 3 + 4 ) + ... + ( 2019 - 2020 )( 2019 + 2020 )

= (-1).3 + (-1).7 + ... + (-1).4039

= -3 - 7 - ... - 4039

= -( 3 + 7 + ... + 4039 )

= \(-\left(\frac{\left(4039+3\right)\left[\left(4039-3\right):4+1\right]}{2}\right)\)

= -2 041 210

Ta có :

B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)

B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\)  (1)

Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\)   (2)

Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)

Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)

\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)

\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)

Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)

Giải:

Ta có:

\(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\) 

\(B=1+\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)\) 

\(B=\dfrac{2021}{2021}+\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}\) 

\(B=2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\) 

\(\Rightarrow\dfrac{A}{B}=\dfrac{\left[2021.\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+...+\dfrac{1}{2}\right)\right]}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=2021\) 

Vậy \(\dfrac{A}{B}=2021\)