Question

Tính nhanh \(1+2^1+2^2+...+2^{100}\)

 

Answer 1

Đặt \(A=1+2^1+2^2+....+2^{100}\)

=>\(2A=2^1+2^2+...+2^{101}\)

=>\(2A-A=\left(2^1+2^2+...+2^{101}\right)-\left(1+2^1+2^2+...+2^{100}\right)\)

=>\(A=2^{101}-1\)

Answer 2

\(A=1+2^1+2^2+...+2^{100}\)

\(2A=2^1+2^2+...+2^{100}+2^{101}\)

\(A=1+2^1+2^2+...+2^{100}\)

\(A=\frac{2^{101}-1}{1}\)

Answer 3

\(A=1+2^1+2^2+...+2^{100}\Rightarrow2A=2^1+2^2+...+2^{100}+2^{101}\)

\(2A-A=2^1+2^2+...+2^{100}+2^{101}-\left(1+2^1+2^2+...+2^{100}\right)\)

\(A=2^{101}-1\)