Câu hỏi của Nguyễn Thị Lê Mi

Question

$Cho$$M_{\left(x\right)}=x+x^2+x^3+...+x^{2016}$a) tìm $M_{\left(3;-3\right)}$b)CM $M⋮105\Leftrightarrow x=2$c)CM $M⋮196\Leftrightarrow x=3$

Phùng Minh Quân · Accepted Answer

$a)$ $M_{\left(3\right)}=3+3^2+3^3+...+3^{2016}$$3M_{\left(3\right)}=3^2+3^3+3^4+...+3^{2017}$$3M_{\left(3\right)}-M_{\left(3\right)}=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)$$2M_{\left(3\right)}=3^{2017}-3$$M_{\left(3\right)}=\frac{3^{2017}-3}{2}$Vậy $M_{\left(3\right)}=\frac{3^{2017}-3}{2}$$M_{\left(-3\right)}=\left(-3\right)+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2016}$$\left(-3\right)M_{\left(-3\right)}=\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+...+\left(-3\right)^{2017}$$\left(-3\right)M_{\left(-3\right)}-M_{\left(-3\right)}=\left[\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2017}\right]-\left[\left(-3\right)+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]$$\left(-4\right)M_{\left(-3\right)}=\left(-3\right)^{2017}+3$$M_{\left(-3\right)}=\frac{\left(-3\right)^{2017}+3}{-4}$$M_{\left(-3\right)}=\frac{-\left(3^{2017}-3\right)}{-4}$$M_{\left(-3\right)}=\frac{3^{2017}-3}{4}$Vậy $M_{\left(-3\right)}=\frac{3^{2017}-3}{4}$Chúc bạn học tốt ~ Câu trả lời: $a)$ $M_{\left(3\right)}=3+3^2+3^3+...+3^{2016}$$3M_{\left(3\right)}=3^2+3^3+3^4+...+3^{2017}$$3M_{\left(3\right)}-M_{\left(3\right)}=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)$$2M_{\left(3\right)}=3^{2017}-3$$M_{\left(3\right)}=\frac{3^{2017}-3}{2}$Vậy $M_{\left(3\right)}=\frac{3^{2017}-3}{2}$$M_{\left(-3\right)}=\left(-3\right)+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2016}$$\left(-3\right)M_{\left(-3\right)}=\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+...+\left(-3\right)^{2017}$$\left(-3\right)M_{\left(-3\right)}-M_{\left(-3\right)}=\left[\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2017}\right]-\left[\left(-3\right)+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]$$\left(-4\right)M_{\left(-3\right)}=\left(-3\right)^{2017}+3$$M_{\left(-3\right)}=\frac{\left(-3\right)^{2017}+3}{-4}$$M_{\left(-3\right)}=\frac{-\left(3^{2017}-3\right)}{-4}$$M_{\left(-3\right)}=\frac{3^{2017}-3}{4}$Vậy $M_{\left(-3\right)}=\frac{3^{2017}-3}{4}$Chúc bạn học tốt ~

Phùng Minh Quân · Answer

$b)$ Ta có : $M_{\left(2\right)}=2+2^2+2^3+...+2^{2016}$$M_{\left(2\right)}=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)$$M_{\left(2\right)}=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2014}\left(1+2+2^2\right)$$M_{\left(2\right)}=2.7+2^4.7+...+2^{2014}.7$$M_{\left(2\right)}=7\left(2+2^4+...+2^{2014}\right)⋮7$ $\left(1\right)$Lại có : $M_{\left(2\right)}=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)$$M_{\left(2\right)}=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{2013}\left(1+2+2^2+2^3\right)$$M_{\left(2\right)}=2.15+2^5.15+...+2^{2013}.15$$M_{\left(2\right)}=15\left(2+2^5+...+2^{2013}\right)⋮15$ $\left(2\right)$Từ (1) và (2) suy ra $M_{\left(2\right)}$ chia hết cho $7$ và $15$$\Rightarrow$$M_{\left(2\right)}⋮105$ ( vì $7.15=105$ ) Vậy nếu $M⋮105$$\Leftrightarrow$$x=2$Chúc bạn học tốt ~ Câu trả lời: $b)$ Ta có : $M_{\left(2\right)}=2+2^2+2^3+...+2^{2016}$$M_{\left(2\right)}=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)$$M_{\left(2\right)}=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2014}\left(1+2+2^2\right)$$M_{\left(2\right)}=2.7+2^4.7+...+2^{2014}.7$$M_{\left(2\right)}=7\left(2+2^4+...+2^{2014}\right)⋮7$ $\left(1\right)$Lại có : $M_{\left(2\right)}=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)$$M_{\left(2\right)}=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{2013}\left(1+2+2^2+2^3\right)$$M_{\left(2\right)}=2.15+2^5.15+...+2^{2013}.15$$M_{\left(2\right)}=15\left(2+2^5+...+2^{2013}\right)⋮15$ $\left(2\right)$Từ (1) và (2) suy ra $M_{\left(2\right)}$ chia hết cho $7$ và $15$$\Rightarrow$$M_{\left(2\right)}⋮105$ ( vì $7.15=105$ ) Vậy nếu $M⋮105$$\Leftrightarrow$$x=2$Chúc bạn học tốt ~

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