Question

[Hoàng Hà Linh]

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

Answer 1

[phạm văn tuấn]

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

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

\(\Rightarrow2B=2^{2017}+2^{2016}+....+2^1+2^0\)

\(\Rightarrow2B-B=(2^{2017}+2^{2016}+...+2^0)-(2^{2016}+2^{2015}+...+2^1+2^0)\)

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

\(\Rightarrow A=2^{2017}-(2^{2017}-1)\)

\(\Rightarrow A=1\)

Answer 2

[Mai Hoàng Giang]

2A = 2²⁰¹⁸ - (2²⁰¹⁷ + 2²⁰¹⁶ + ....+ 2¹⁾

2A - A = [2²⁰¹⁸ - (2²⁰¹⁷ + 2²⁰¹⁶ + ....+ 2¹ )] - [2^{2017 -} (2²⁰¹⁶ + 2²⁰¹⁵ +..... + 2¹ + 2⁰)

A = 2²⁰¹⁸  -  2²⁰¹⁷ - 2²⁰¹⁷ - 1 

A = 2²⁰¹⁸ - (2²⁰¹⁷ +2²⁰¹⁷ +1)

A = 2²⁰¹⁸ - (2²⁰¹⁸ +1 )

A = -1