Question

 Cho \(N=2^{2019}-2^{2018}-2^{2017}-...-2-1\)

Tính\(A=20^n+11^n+2019^n\)

Answer 1

\(N=2^{2019}-2^{2018}-2^{2017}-...-2-1\)

   \(=2^{2019}-\left(2^{2018}+2^{2017}+...+2+1\right)\)

Đặt   \(B=1+2+...+2^{2017}+2^{2018}\)

\(\Rightarrow\) \(2B=2+2^2+...+2^{2018}+2^{2019}\)

\(\Rightarrow\) \(B=2^{2019}-1\)

\(\Rightarrow\) \(N=2^{2019}-2^{2019}+1=1\)

\(\Rightarrow\) \(A=20^1+11^1+2019^1\)

            \(=20+11+2019\)

           \(=2050\)

Study well ! >_<

Answer 2

N=\(2^{2019}-\left(1+2+.....2^{2018}\right)\)

Đặt B=\(1+2+..........+2^{2018}\)

2B=\(2+2^2+..........+2^{2019}\)

2B-B=B=\(2^{2019}-1\)

Suy ra N=\(2^{2019}-2^{2019}+1=1\)

A=20+11+2019=2050

hok tốt