Question

\(2^1+2^2+2^3+...+2^{50}\)

Answer 1

\(2^1+2^2+2^3+...+2^{50}\\ =\left(2-1\right)\left(2^1+2^2+2^3+...+2^{50}\right)\\ =2^2-2+2^3-2^2+2^4-2^3+...+2^{51}-2^{50}\\ =2^{51}-2\)

Answer 2

ta đặc \(2^1+2^2+2^3+...+2^{50}\) là \(B\)

\(\Rightarrow2B=2\left(2^1+2^2+2^3+...+2^{50}\right)\)

\(2B=2^2+2^3+2^4+...+2^{51}\)

ta có : \(2B-B=B=\left(2^2+2^3+2^4+...+2^{51}\right)-\left(2^1+2^2+2^3+...+2^{50}\right)\)

\(B=2^{51}-2^1=2^{51}-2\)

vậy \(2^1+2^2+2^3+...+2^{50}=2^{51}-2\)

Answer 3

dsfsffsdfsdf

Answer 4

Đặt \(A=2^1+2^2+2^3+.....+2^{50}\)

\(\Rightarrow2A=2\left(2^1+2^2+2^3+....+2^{50}\right)\)

\(\Rightarrow2A=2^2+2^3+2^4+....+2^{51}\)

\(\Rightarrow2A-A=\left(2^2+2^3+2^4+....+2^{51}\right)-\left(2^1+2^2+2^3+....+2^{50}\right)\)

\(\Rightarrow A=2^{51}-2\)