Ta có: 

2^n -1-2-2^2-2^3- ......... - 2^100 = 1

=> 2^n= 1+1+2+2^2+2^3+ ........ + 2^100.

=> 2 x 2^n= 2+2+4+2^3+2^4+ ....... + 2^101

=> 2^n = 2 x 2^n - 2^n= (2+2+4+2^3+2^4+......+2^101) - (1+1+2+2^2+2^3+ ....... + 2^100) =(2 + 2^101) - ( 1+1)= 2 + 2^101 - 2 = 2^101.

=> n= 101.

Ta có 2A=\(2^2+2^3+...+2^{101}\)

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

=> A= \(2^{101}-2\)

Mà \(A+1=2^x\)

=> \(2^x=2^{101}-2^0\)

Bạn xem lại đề nhé mk cx ko rõ nữa 

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

2A=\(2^2+2^3+2^4+.....+2^{101}\)

\(2A-A=\left(2^2+2^3+2^4+...2^{101}\right)-\left(2+2^2+2^3+....+2^{100}\right)\)

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

Vậy A= \(2^{101}-2\)

bn hiếu cho mk hỏi x bằng bao nhiêu vậy

( x + 1 ) + ( x + 2 ) + ( x + 3 ) + .... + ( x + 100 ) = 5750

( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750

     100 chữ số x

100 . x + 5050 = 5750

100 . x = 5750 - 5050

100 . x = 700

x = 700 : 100

x = 7

(x+x+...+x)+(1+2+3+...+100)=5750

100x+5050=5750

100x=5750-5050

100x=700

x=700/100

x=7

2n+5chia hết cho 2n+1

=>4n+10chia hết cho 4n+2

=>2n+5chia hết cho 2n+1

Ta có: 2n + 5 = (2n - 1) + 6

Do 2n - 1 \(⋮\)2n - 1 => 6 \(⋮\)2n - 1

=> 2n - 1 \(\in\)Ư(6) = {1; 2; 3; 6}

=> 2n \(\in\){2; 3; 4; 7}

Do n \(\in\)N=> n \(\in\){1; 2}

Mình k cho bạn Edogawa Cona rùi nhé.Thanks

1/ \(=3^n.3^2+3^n=3^n\left(3^2+1\right)=10.3^n⋮10\)

2/ \(100.x+\left(1+2+3+...+100\right)=7450\)

Đến đây bạn tự làm nốt nhé

1. Ta có: \(3^{n+2}+3^n=3^n.\left(3^2+1\right)=3^n.\left(9+1\right)=3^n.10⋮10\)( đpcm )

2. \(\left(x+1\right)+\left(x+2\right)+.......+\left(x+100\right)=7450\)

\(\Leftrightarrow x+1+x+2+........+x+100=7450\)

\(\Leftrightarrow100x+\frac{100.101}{2}=7450\)

\(\Leftrightarrow100x+5050=7450\)

\(\Leftrightarrow100x=2400\)\(\Leftrightarrow x=24\)

Vậy \(x=24\)

1. 3ⁿ⁺² + 3ⁿ = 3ⁿ( 3² + 1 ) = 3ⁿ.10 \(⋮\)10 \(\forall n\inℕ\)( đpcm )

2. ( x + 1 ) + ( x + 2 ) + ... + ( x + 100 ) = 7450

<=> ( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 7450

<=> 100x + \(\frac{\left(100+1\right)\left[\left(100-1\right):1+1\right]}{2}\)= 7450

<=> 100x + 5050 = 7450

<=> 100x = 2400

<=> x = 24

Bài 2 :

n + 5 chia hết cho n

=> 5 chia hết cho n

=> n thuộc Ư(5) = {1 ; 5}

b) 2016.(n - 3) + 11 chia hết cho n - 3

=> 11  chia hết cho n - 3

=> n - 3 thuộc Ư(11) = {1 ; 11}\

=> n = {4 ; 14}

c) n² + 2n + 3 chia hết cho n + 2

n.(n + 2) + 3 chia hết cho n + 2

=> 3 chia hết cho n + 2

=> n + 2 thuộc U(3) = {1 ; 3}

=> n = {-1 ; 1}

a) 2(x + 2) + 3x = 29

2x + 4 + 3x = 29

5x = 29 - 4 = 25

x = 5

b) 720:[41 - (2x-5)]=2³ . 5

41 - (2x - 5) = 720 : 40 = 180

2x - 5 = 41 - 180 = -139

2x = -139 + 5 = -134

x = (-134) : 2 = -67

c) (x + 1) + (x + 2) + ..... + (x + 100) = 5750

x + 1 + x + 2 + ........ + x + 100 = 5750

100x + (1 + 2 + 3 + ........... + 100) = 5750

100x + 5050 = 5750

100x = 700

x = 7