Câu hỏi của Nguyễn Diệu Anh

Question

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

Trần Đặng Phan Vũ · Accepted Answer

$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}$Câu trả lời: $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}$

Phùng Minh Quân · Answer

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 $A=\frac{2^{2016}+2}{3}$Câu trả lời: 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 $A=\frac{2^{2016}+2}{3}$

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