Question

[Lực Võ]

Tính

A=2^2017-(2^2016+2^2015+...+2^2+2^1+2^0)

Answer 1

[tth_new]

Theo mình bạn không nên hỏi những bài thế này vì nó rất easy

\(A=2^{2017}-\left(2^{2016}+2^{2015}+...+2^2+2^1+2^0\right)\)

\(A=2^{2017}-\left(2^0+2^1+2^2+...+2^{2015}+2^{2016}\right)\)

\(B=\left(2^0+2^1+2^2+...+2^{2015}+2^{2016}\right)\)

\(2B=\left(2^1+2^2+2^3+...+2^{2016}+2^{2017}\right)\)

\(B=\left(2^1+2^2+2^3+...+2^{2016}+2^{2017}\right)-\left(2^0+2^2+2^3+...+2^{2016}+2^{2017}\right)\)

\(B=2^{2017}-2^0=2^{2017}-1\)

\(A=2^{2017}-\left(2^{2017}-1\right)=2^{2017}-2^{2017}+1=1\)