Question

[Nguyễn Diệu Anh]

Tính nhanh: 

A = 2 - 2² + 2³ - 2⁴ + 2⁵ - 2⁶ + ... + 2²⁰¹⁵

Answer 1

[Trần Đặng Phan Vũ]

\(A=2-2^2+2^3-2^4+......+2^{2015}\)

\(2A=2^2-2^3+2^4-2^5+.....+2^{2016}\)

\(2A+A=2^2-2^3+2^4-2^5+.....+2^{2016}+\left(2-2^2+2^3-2^4+.....+2^{2015}\right)\)

\(3A=2^{2016}+2\)

\(\Rightarrow A=\frac{2^{2016}+2}{3}\)

Answer 2

[Phùng Minh Quân]

Ta có : 

\(A=2-2^2+2^3-2^4+...+2^{2015}\)

\(\Leftrightarrow\)\(A=\left(-2^2-2^4-...-2^{2014}\right)+\left(2+2^3+...+2^{2015}\right)\)

\(\Leftrightarrow\)\(A=-\left(2^2+2^4+...+2^{2014}\right)+\left(2+2^3+...+2^{2015}\right)\)

Gọi \(M=2^2+2^4+...+2^{2014}\)

\(\Leftrightarrow\)\(4M=2^4+2^6+...+2^{2016}\)

\(\Leftrightarrow\)\(4M-M=\left(2^4+2^6+...+2^{2016}\right)-\left(2^2+2^4+...+2^{2014}\right)\)

\(\Leftrightarrow\)\(3M=2^{2016}-2^2\)

\(\Leftrightarrow\)\(M=\frac{2^{2016}-4}{3}\)

Gọi \(N=2+2^3+...+2^{2015}\)

\(\Leftrightarrow\)\(4N=2^3+2^5+...+2^{2017}\)

\(\Leftrightarrow\)\(4N-N=\left(2^3+2^5+...+2^{2017}\right)-\left(2+2^3+...+2^{2015}\right)\)

\(\Leftrightarrow\)\(3N=2^{2017}-2\)

\(\Leftrightarrow\)\(N=\frac{2^{2017}-2}{3}\)

\(\Rightarrow\)\(A=-\left(2^2+2^4+...+2^{2014}\right)+\left(2+2^3+...+2^{2015}\right)=-\frac{2^{2016}-4}{3}+\frac{2^{2017}-2}{3}\)

\(\Rightarrow\)\(A=\frac{\left(-1\right).\left(2^{2016}\right)+2^{2017}.1+4-2}{3}=\frac{2^{2016}\left(2-1\right)+2}{3}=\frac{2^{2016}+2}{3}\)

Vậy \(A=\frac{2^{2016}+2}{3}\)