Question

[Hải Đăng Chu Lê]

Tìm số nguyên x biết: 2^x+2^x+1+2^x+2+…+2^x+2020=2²⁰²⁴-8

Answer 1

[Akai Haruma]

Lời giải:

$2^x+2^{x+1}+2^{x+2}+...+2^{x+2020}=2^{2024}-8$

$2^x(1+2+2^2+...+2^{2020})=2^{2024}-8(1)$

$2^x(2+2^2+2^3+...+2^{2021})=2^{2025}-16(2)$

Lấy $(2)$ trừ $(1)$ ta có:

$2^x(2^{2021}-1)=2^{2025}-16-(2^{2024}-8)=2^{2024}(2-1)-8$

$2^x(2^{2021}-1)=2^{2024}-8=2^3(2^{2021}-1)$

$\Rightarrow 2^x=2^3$
$\Rightarrow x=3$