\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)

B=2+22+23+24+...+299+2100=2(1+22+23+24)+...+296(1+22+23+24)=2.31+26.31+...+296.31=31(2+26+...+296)⋮31

\(=\dfrac{20}{21}x\dfrac{21}{22}x\dfrac{22}{23}x...x\dfrac{1999}{2000}\)

\(=\dfrac{20}{2000}=\dfrac{1}{100}\)

=20/21x21/22x22/23x..............x1998/1999x1999/2000

=20x21x22x23x.....................x1998x1999/21x22x23x24x...............x1999x2000

=20/2000

1/100

A = (1- \(\dfrac{1}{21}\))\(\times\)( 1 - \(\dfrac{1}{22}\))\(\times\)(1-\(\dfrac{1}{23}\))\(\times\)...\(\times\)(1-\(\dfrac{1}{1999}\))\(\times\)(1- \(\dfrac{1}{2000}\))

A = \(\dfrac{21-1}{21}\)\(\times\)\(\dfrac{22-1}{22}\)\(\times\)\(\dfrac{22-1}{23}\)\(\times\)...\(\times\)\(\dfrac{1999-1}{1999}\)\(\times\)\(\dfrac{2000-1}{2000}\)

A = \(\dfrac{20}{21}\)\(\times\)\(\dfrac{21}{22}\)\(\times\)\(\dfrac{1998}{1999}\)\(\times\)\(\dfrac{1999}{2000}\)

A = \(\dfrac{21\times....\times1999}{21\times...\times1999}\) \(\times\) \(\dfrac{20}{2000}\)

A= \(\dfrac{1}{100}\)

\(2\cdot A=2^2+2^3+...+2^{101}\)

\(\Leftrightarrow A=2^{101}-2\)

Đặt \(A=2+2^2+...+2^{100}\)

\(2A=2^2+2^3+...+2^{101}\\ 2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\\ A=2^{101}-2\)

\(A=1\cdot2+2\cdot3+...+151\cdot152\)

\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)

\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)

\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)

\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)

\(=151\cdot76+151\cdot7676=1170552\)

\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)

\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)

\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)

\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)

\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)

\(=4\left[506\cdot1013+345990150\right]\)

\(=1386010912\)

\(M=1^2+2^2+...+2024^2\)

\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)

\(=2024\cdot2025\cdot\dfrac{4049}{6}\)

=2765871900

\(N=1^3+2^3+...+100^3\)

\(=\left(1+2+3+...+100\right)^2\)

\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)

\(=\left[50\cdot101\right]^2=5050^2\)

\(Q=1^3+2^3+...+2024^3\)

\(=\left(1+2+3+...+2024\right)^2\)

\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)

\(=\left[1012\left(2024+1\right)\right]^2\)

\(=2049300^2\)

41 + 42 + 43 + 44 - 21 - 22 - 23 - 24

= (41 - 21) + (42 - 22) + (43 - 23) + (44 - 24)

= 20 + 20 + 20 + 20

= 20 x 4

= 80

tk nha. chúc bn học giỏi :)

= 80 nha

TK MK NHA CÁC BẠN MK ĐG BỊ ÂM ĐIỂM

Giúp em thì ai giúp anh, chị đây! Hỏi mãi chưa có chế nào trả lời đây. Huhuhu$$$

a) Ta có A = 2¹ + 2² + 2³ + ... + 2²⁰²²

2A = 2² + 2³ + 2⁴ + ... + 2²⁰²³

2A - A = ( 2² + 2³ + 2⁴ + ... + 2²⁰²³ ) - ( 2¹ + 2² + 2³ + ... + 2²⁰²² )

A = 2²⁰²³ - 2

Lại có B = 5 + 5² + 5³ + ... + 5²⁰²²

5B = 5² + 5³ + 5⁴ + ... + 5²⁰²³

5B - B = ( 5² + 5³ + 5⁴ + ... + 5²⁰²³ ) - ( 5 + 5² + 5³ + ... + 5²⁰²² )

4B = 5²⁰²³ - 5

B = \(\dfrac{5^{2023}-5}{4}\)

b) Ta có : A + 2 = 2^x

⇒ 2²⁰²³ - 2 + 2 = 2^x

⇒ 2²⁰²³ = 2^x

Vậy x = 2023

Lại có : 4B + 5 = 5^x

⇒ 4 . \(\dfrac{5^{2023}-5}{4}\) + 5 = 5^x

⇒ 5²⁰²³ - 5 + 5 = 5^x

⇒ 5²⁰²³ = 5^x

Vậy x = 2023

bạn viết rõ lũy thừa giúp mình với

\(A=B\)

ý bạn là  như này đk?

A=19²¹+1:19²²+1

B=19²²+1:19²³+1