Lời giải:

$2^n+34=2.2^2+3.2^3+....+n.2^n$

$2^{n+1}+68=2.2^3+3.2^4+....+n.2^{n+1}$

Trừ theo vế:

$2^n+34=n.2^{n+1}-(8+2^3+2^4+...+2^n)$

$n.2^{n+1}-2^n-42=2^3+2^4+...+2^n$

$n.2^{n+2}-2^{n+1}-84=2^4+....+2^{n+1}$

Trừ theo vế:

$n.2^{n+1}-2^n-42=2^{n+1}-8$

$2^n(2n-3)=34=17.2$

$\Rightarrow 2^n=2$ và $2n-3=17$ (vô lý)

Vậy không tìm được $n$.

A = 2.2² + 3.2³ + 4.2⁴ + ... + n.2ⁿ 

2.A = 2.2³ + 3.2⁴ + 4.2⁵ + ...+ n.2ⁿ⁺¹

=> A - 2.A = 2.2² + (3.2³ - 2.2³)  + (4.2⁴ - 3.2⁴) + ...+ (n - n + 1).2ⁿ - n.2ⁿ⁺¹

=> A = 2.2² + 2³ + 2⁴ + ..+ 2ⁿ - n.2^{n+ 1}  = 2² + (2² + 2³ + ....+ 2^{n+ 1}) - (n+1).2ⁿ⁺¹

=> A =  - 2² -  (2² + 2³ + ....+ 2^{n+ 1}) + (n+1).2ⁿ⁺¹

Tính B = 2² + 2³ + ....+ 2^{n+ 1} => 2.B =  2³ + ....+ 2^{n+ 1} + 2ⁿ⁺² => 2B - B = 2ⁿ⁺² - 2² => B = 2ⁿ⁺² - 2²

Vậy A = 2² - 2ⁿ⁺² + 2² + (n+1).2ⁿ⁺¹ = (n+1).2ⁿ⁺¹ - 2^{n+ 2} = 2ⁿ⁺¹.(n + 1 - 2) = (n-1).2ⁿ⁺¹ = 2(n-1).2ⁿ

Theo bài cho  A = 2(n-1).2ⁿ = 2ⁿ⁺¹⁰ => 2(n - 1) = 2¹⁰ => n - 1 = 2⁹  = 512 => n = 513

Vậy.............

n= 513, tui chỉ biết đáp án nhưng không biết cách làm

đặt A=2+2^2+2^3+...+2^n

     2A=2^2+2^3+2^4+...+2^n+1 (1)

  2A-A=2\(^{n+1}\)-2

     A=2\(^{n+1}\)-2  (2)

từ (1)(2) =>2 + 2\(^2\)+2\(^3\)+...+2\(^n\)=2\(^{n+1}\)-2

                      2\(^2\)+2\(^3\)+...+2\(^n\)=2\(^{n-1}\)-2\(^2\)

                              ..............................

                                             2\(^n\)=2\(^{n-1}\)-2\(^n\)

cộng vế với vế ta có 

 2+2.2\(^2\)+3.2\(^3\)+...+n.2\(^n\)= n.2\(^{n+1}\)- (2+2\(^2\)+2\(^3\)+...+2\(^n\))

2+(2.2\(^2\)+3.2\(^3\)+...+n.2\(^n\)=n.2\(^{n+1}\)- A

     2+2\(^{n+10}\)=n.2\(^{n+1}\)-2\(^{n+1}\)+2

            2\(^{n+10}\)=2\(^{n+1}\).(n-1)

             2\(^{n+1}\). 2\(^9\)=2\(^{n+1}\).(n-1)

=>n-1=2\(^9\)

=>n=2^9+1=513

vậy n=513

a, 5ⁿ⁺¹ - 5^n-1 = 125⁴.2³.3

5^n-1.(5² - 1) = 125⁴.24

5^n-1.24         = 125⁴.24

5^n-1             = 125⁴

5^n-1             = (5³)⁴

5^n-1            = 5¹²

n - 1           = 12

n                = 12 + 1

n                = 13

b,2^2n-1 + 2²ⁿ⁺² = 3.2¹¹

   2^2n-1.(1 + 2³) = 3.2¹¹

  2^2n-1.9 = 3.2¹¹

 2^2n-1      = 2¹¹: 3

2²ⁿ        = 2¹² : 3 (xem lại đề bài em nhá)

Đặt S = 2.2² + 3.2³ + 4.2⁴ + ... + (n - 1).2^{n - 1} + n.2ⁿ 

<=> S = 2S - S = (2.2³ + 3.2⁴ +  4.2⁵ + .... + (n - 1).2ⁿ + n. 2^{n + 1}) - (2.2² + 3.2³ + 4.2⁴ + ... + (n - 1).2^{n - 1} + n.2ⁿ)

                S = (2.2³ - 3.2³) + (3.2⁴ - 4.2⁴) + (4.2⁵ - 5.2⁵) + .... + [(n - 1).2ⁿ - n.2ⁿ] + n.2^{n + 1} - 2.2²

                   = -(2³ + 2⁴ + 2⁵ + ... + 2ⁿ) + n.2^{n + 1} - 8

Đặt A = 2³ + 2⁴ + 2⁵ + ... + 2ⁿ

  <=> 2A - A = (2⁴ + 2⁵ + 2⁶ + ... + 2^{n + 1}) - (2³ + 2⁴ + 2⁵ + ... + 2ⁿ) 

  <=> A = 2^{n + 1} - 2³ 

Khi đó S = - 2^{n - 1} + 2³ + n.2^{n - 1} - 8

              = 2^{n - 1}.(n - 1) = 2^{n + 34}

         => n - 1 = 2^{n + 34} : 2^{n - 1}

          => n - 1 = 2^{n + 34 - n + 1}

          => n - 1 = 2³⁵

          => n = 2³⁵ + 1

N=34359738369 nha

Cường xo tính lại kết quả của Xyz mà cũng làm

a) \(5^{n+3}-5^{n+1}=5^{12}.120\Leftrightarrow5^{n+1}.\left(5^2-1\right)=5^{12}.5.24\)

\(\Leftrightarrow24.5^{n+1}=5^{13}.24\Leftrightarrow5^{n+1}=5^{13}\Leftrightarrow n+1=13\Leftrightarrow n=12\)

b) \(2^{n+1}+4.2^n=3.2^7\)

\(\Leftrightarrow2^n\left(2+4\right)=3.2^7\Leftrightarrow6.2^n=3.2^7\Leftrightarrow2^n=2^6\Leftrightarrow n=6\)

c) \(3^{n+2}-3^{n+1}=486\)

\(\Leftrightarrow3^{n+1}.\left(3-1\right)=486\Leftrightarrow2.3^{n+1}=486\Leftrightarrow3^{n+1}=243\)

\(\Leftrightarrow3^n=243:3=81=3^3\Leftrightarrow n=3\)

d) \(3^{2n+3}-3^{2n+2}=2.3^{10}\)

\(\Leftrightarrow3^{2n+2}.\left(3-1\right)=2.3^{10}\)

\(\Leftrightarrow3^{2n+2}.2=2.3^{10}\Leftrightarrow3^{2n+2}=3^{10}\Leftrightarrow2n+2=10\Leftrightarrow2n=8\Leftrightarrow n=4\)

Xét khai triển:

\(\left(1+2x\right)^{2n+1}=C_{2n+1}^0+C_{2n+1}^1.2x+C_{2n+1}^2\left(2x\right)^2+...+C_{2n+1}^{2n+1}\left(2x\right)^{2n+1}\)

Đạo hàm 2 vế:

\(2\left(2n+1\right)\left(1+2x\right)^{2n}=2C_{2n+1}^1+2^2C_{2n+1}^2x+...+\left(2n+1\right)2^{2n+1}C_{2n+1}^{2n+1}x^{2n}\)

\(\Leftrightarrow\left(2n+1\right)\left(1+2x\right)^{2n}=C_{2n+1}^1+2C_{2n+1}^2x+...+\left(2n+1\right)2^{2n}C_{2n+1}^{2n+1}x^{2n}\)

Cho \(x=-1\) ta được:

\(2n+1=C_{2n+1}^1-2C_{2n+1}^2+...+\left(2n+1\right)2^{2n}C_{2n+1}^{2n+1}\)

\(\Rightarrow2n+1=2019\Rightarrow n=1009\)