Question

2^x+2^x+1+2^x+2+...+2^2018=2^2021-4

Answer 1

Ta có: \(2^{x}+2^{x+1}+\cdots+2^{x+2018}=2^{2021}-4\)

=>\(2^{x}\left(1+2+\cdots+2^{2018}\right)=2^2\left(2^{2019}-1\right)\)

Đặt \(A=1+2+\cdots+2^{2018}\)

=>\(2A=2+2^2+2^3+\ldots+2^{2019}\)

=>\(2A-A=2^2+2^3+\cdots+2^{2019}-1-2-\cdots-2^{2018}\)

=>\(A=2^{2019}-1\)

Ta có: \(2^{x}\left(1+2+\cdots+2^{2018}\right)=2^2\left(2^{2019}-1\right)\)

=>\(2^{x}\left(2^{2019}-1\right)=2^2\left(2^{2019}-1\right)\)

=>\(2^{x}=2^2\)

=>x=2