Question

[Lê Tùng Anh]

Tìm n ϵ N: 2.$2^2$ + 3.$2^3$ + 4.$2^4$ +...+ n.$2^n$ = $2^{n+11}$

Answer 1

[Akai Haruma]

Lời giải:

Đặt $A=2.2^2+3.2^3+4.2^4+...+n.2^n$

$2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}$

$\Rightarrow A=2A-A=n.2^{n+1}-(2^3+2^4+...+2^n) - 2.2^2$

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

Đặt $S=2^3+2^4+...+2^n$

$2S=2^4+2^5+...+2^{n+1}$

$\Rightarrow S=2S-S=2^{n+1}-2^3=2^{n+1}-8$

$\Rightarrow A=n.2^{n+1}-S-8 = n.2^{n+1}-2^{n+1}+8-8=(n-1).2^{n+1}$

Vậy $(n-1).2^{n+1}=2^{n+11}$

$\Rightarrow n-1 = 2^{10}\Rightarrow n=2^{10}+1=1025$