Question

[Minhchau Trần]

Cmr (2+1) (2^2+1) (2^4+1)...(2^1024+1)=1+2+2^2+...+2^2047

Answer 1

[Akai Haruma]

Lời giải:

$\text{VT}=(2^2-1)(2^2+1)(2^4+1)...(2^{1024}+1)$

$=(2^4-1)(2^4+1)....(2^{1024}+1)$

$=(2^8-1)(2^8+1)....(2^{1024}+1)$

$=(2^{1024})^2-1=2^{2048}-1$

$\text{VP}=1+2+...+2^{2047}$

$2\text{VP}=2+2^2+...+2^{2048}$

$\Rightarrow 2\text{VP}-\text{VP}=2^{2048}-1$

$\Leftrightarrow \text{VP}=2^{2048}-1$

Vậy $\text{VT}=\text{VP}$

Answer 2

[Trên con đường thành công không có....]