Question

Số dư của \(A=2^0+2^1+2^2+...+2^{2016}\) khi chia cho 8 là ....

Answer 1

Co :

\(A=1+2^1+2^2+......+2^{2016}\)

\(\Rightarrow2A=2^1+2^2+2^3+.....+2^{2016}+2^{2017}\)

ma A=2A-A=(\(2^1+2^2+2^3+......+2^{2016}+2^{2017}\))-(\(1+2^1+2^2+2^3+....+2^{2016}\))

\(\Rightarrow A=2^{2016}-1\)

\(\frac{2^{2016}}{8}\in N\)

Vay so du cua A khi chia cho 8 la 7

nhan tien cho minh hoi luon bao gio thi Ha Nam thi violympic cap huyen day?Cam on ban nha