Câu hỏi của arthur

Question

C/M 1+2+22+....+299+2100=2101-1

Đan Anh · Accepted Answer

Đặt $A=1+2+2^2+...+2^{99}+2^{100}$

$\Rightarrow2A=2.\left(1+2+2^2+...+2^{99}+2^{100}\right)$

$=2+2^2+2^3+...+2^{100}+2^{101}$

Có $2A-A=\left(2+2^2+2^3+...+2^{100}+2^{101}\right)-\left(1+2^{ }+2^2+...+2^{99}+2^{100}\right)$

$A=2+2^2+2^3+...+2^{100}+2^{101}-1-2-2^2-...-2^{99}-2^{100}$

$A=2^{101}-1$ (đpcm)Câu trả lời: Đặt $A=1+2+2^2+...+2^{99}+2^{100}$

$\Rightarrow2A=2.\left(1+2+2^2+...+2^{99}+2^{100}\right)$

$=2+2^2+2^3+...+2^{100}+2^{101}$

Có $2A-A=\left(2+2^2+2^3+...+2^{100}+2^{101}\right)-\left(1+2^{ }+2^2+...+2^{99}+2^{100}\right)$

$A=2+2^2+2^3+...+2^{100}+2^{101}-1-2-2^2-...-2^{99}-2^{100}$

$A=2^{101}-1$ (đpcm)

Violympic toán 8