giúp mik nha

\(\Leftrightarrow9A=3^3+3^5+...+3^{21}\\ \Leftrightarrow9A-A=3^3+3^5+...+3^{21}-3-3^3-3^5-...-3^{19}\\ \Leftrightarrow8A=3^{21}-3\Leftrightarrow A=\dfrac{3^{21}-3}{8}\)

A=1-2+3-4+5-6+.....+99-100+101

A = (1  - 2 ) + ( 3 - 4 ) + ( 5 - 6 ) + ... + ( 99 - 100 ) + 101

A = ( -1 ) + ( -1 ) + ( -1 ) + ... + ( -1 ) + 101

A = ( -1 ) . 50 + 101

A = -50 + 101

A = 51

a: \(2A=2^1+2^2+...+2^{2022}\)

\(\Leftrightarrow A=2^{2022}-1\)

\(A=1+2+2^2+...+2^{2021}\)

\(2A=2+2^2+2^3+...+2^{2020}\)

\(2A-A=\left(2+2^2+2^3+...+2^{2020}\right)-\left(1+2+2^2+...+2^{2021}\right)\)

\(A=2^{2020}-1\)

A=2+2^3+2^5+...+2^2009

4A=2^3+2^5+2^7+...+2^2011

4A-A=(2^3+2^5+2^7+...+2^2011)-(2+2^3+2^5+...+2^2009)

3A=2^2011-2

A=(2^2011-2):3

a: 5A=5+5^2+...+5^2023

=>4A=5^2023-1

=>A=(5^2023-1)/4

b: 6B=6^2+6^3+...+6^41

=>5B=6^41-6

=>B=(6^41-6)/5

c: 16C=4^4+4^6+...+4^16

=>15C=4^16-4^2

=>C=(4^16-4^2)/15

d: 9D=3^3+3^5+...+3^27

=>8D=3^27-3

=>D=(3^27-3)/8

A=[(99-3):3+1].(99+3):2=33.102:2=33.51=1683

C=[(99-3):3+1].(99+3):2=33.102:2=33.51=1683

B=[(120-2):2+1].(120+2):2=60.122:2=60.61=3660

A=[(100-1):1+1].(100+1):2=100.101:2=50.101=5050

D=[(123-3):3+1].(123+3):2=41.126:2=41.63=2583

vừa nãy cậu chx đăng hết câu hỏi nên mik làm 1 câu

a: \(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{1}{3}-\dfrac{1}{203}=\dfrac{200}{609}\)

b: \(B=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{76}\)

\(=\dfrac{1}{4}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)

\(A=\dfrac{\left(\dfrac{122-24}{1}+1\right)\left(122+24\right)}{2}=7227\)

\(B=\dfrac{\left(\dfrac{1995-12}{3}+1\right)\left(1995+12\right)}{2}=664317\)

a, 3650

b, 664317

c, 50 

d, 1022121

`#3107.101107`

1.

`a,`

\(A=1+3+3^2+3^3+...+3^{2012}\)

`3A = 3 + 3^2 + 3^3 + ... + 3^2013`

`3A - A = (3 + 3^2 + 3^3 + ... + 3^2013) - (1 + 3 + 3^2 + 3^3 + ... + 3^2012)`

`2A = 3 + 3^2 + 3^3 + ... + 3^2013 - 1 - 3 - 3^2 - 3^3 - ... - 3^2012`

`2A = 3^2013 - 1`

`=> A = (3^2013 - 1)/2`

Vậy, `A = (3^2013 - 1)/2`

`b,`

\(B=1+10+10^2+10^3+...+10^{2023}\)

`10B = 10 + 10^2 + 10^3 + ... + 10^2024`

`10 B - B = (10 + 10^2 + 10^3 + ... + 10^2024) - (1 - 10 + 10^2 + 10^3 + ... + 10^2023)`

`9B = 10 + 10^2 + 10^3 + ... + 10^2024 - 1 - 10^2 - 10^3 - ... - 10^2023`

`9B = 10^2024 - 1`

`=> B = (10^2024 - 1)/9`

Vậy, `B = (10^2024 - 1)/9.`

`a)A=1+3+3^2+3^3+...+3^2012`

`=>3A=3+3^2+3^3+...+3^2013`

`=>3A-A=2A=3^2013-1`

`=>A=(3^2013-1)/2`

`b)B=1+10+10^2+...+10^2024`

`=>10B=10+10^2+10^3+....+10^2025`

`=>10B-B=9B=10^2025-10`

`=>B=(10^2025-10)/9`

a)

`1/1-1/2`

`=2/2-1/2`

`=1/2`

b)

`1/(1*2)+1/(2*3)`

`=1/1-1/2+1/2-1/3`

`=1/1-1/3`

`=3/3-1/3`

`=2/3`

c)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

d) 

\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?

\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)