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.2n+ 1 = 22 + (22 + 23 + ....+ 2n+ 1) - (n+1).2n+1
=> A = - 22 - (22 + 23 + ....+ 2n+ 1) + (n+1).2n+1
Tính B = 22 + 23 + ....+ 2n+ 1 => 2.B = 23 + ....+ 2n+ 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 - 2n+ 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
\(\sqrt{\sqrt[]{}\frac{f}{f}\hept{\begin{cases}s\\\end{cases}}sf}\)
Đặt A = 2 . 22 + 3 . 23 + 4 . 24 + ... + n . 2n
=> 2A = 2 . 23 + 3 . 24 + 4 . 25 + ... + n . 2n+1
=> 2A - A = n . 2n+1 - 2n - ... - 24 - 23 - 2 . 22
=> A = n . 2n+1 - (2n + ... + 24 + 23 + 23)
=> A = n . 2n+1 - (2n + ... + 24 + 24) (vì 2n + 2n = 2n+1)
=> A = n . 2n+1 - 2n+1 = 2n+1(n - 1)
Kết hợp với giả thiết => 2n+1(n - 1) = 2n+10
=> n - 1 = 2n+10 : 2n+1
=> n - 1 = 29 = 512
=> n = 512 + 1 = 513
Vậy n = 513