Question

Tìm E :

E=1+2+2^2+…….+2^2017

Answer 1

\(E=1+2+2^2+…….+2^{2017}\)

\(\Rightarrow2E=2\left(1+2+2^2+2^3+...+2^{2017}\right)\)

\(\Rightarrow2E=2+2^2+2^3+...+2^{2018}\)

\(\Rightarrow2E-E=2+2^2+2^3+...+2^{2018}-1-2-2^2-...-2^{2017}\)

\(\Rightarrow E=2^{2018}-1\)

Vậy \(E=2^{2018}-1\)

Answer 2

Mk nghĩ: E= 2^{2018 -} 1

Answer 3

\(E=1+2+...+2^{2017}\)

\(\Rightarrow2E=2+2^2+...+2^{2018}\)

\(\Rightarrow2E-E=\left(2+2^2+...+2^{2018}\right)-\left(1+2+...+2^{2017}\right)\)

\(\Rightarrow E=2^{2018}-1\)

Answer 4

\(E=1+2+2^2+............+2^{2017}\)

\(\Leftrightarrow2E=2+2^2+..........+2^{2017}+2^{2018}\)

\(\Leftrightarrow2E-E=\left(2+2^2+........+2^{2018}\right)-\left(1+2+.....+2^{2017}\right)\)

\(\Leftrightarrow E=2^{2018}-1\)

Answer 5

E = 1 + 2 + 2² + ... + 2²⁰¹⁷
=> 2E = 2 + 2² + ... + 2²⁰¹⁷ + 2²⁰¹⁸
=> 2E - E = 2²⁰¹⁸ - 2
=> E = 2²⁰¹⁸ - 2
@Yuuki Tenpouin

Answer 6

Các bạn lm sai hết nhé. Người ta bảo tìm E mà các bạn tính E