Question

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

Answer 1

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

Answer 3

Đặt \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1024}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1024}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1024}+1\right)=2^{2048}-1\)

Đặt \(B=1+2+2^2+...+2^{2047}\)

\(\Rightarrow2B=2+2^2+2^3+...+2^{2048}\)

\(\Rightarrow B=2B-B=2+2^2+2^3+...+2^{2048}-1-2-2^2-...-2^{2047}=2^{2048}-1\)