Question

[Nguyễn Thị Lê Mi]

\(Cho\)\(M_{\left(x\right)}=x+x^2+x^3+...+x^{2016}\)

a) tìm \(M_{\left(3;-3\right)}\)

b)CM \(M⋮105\Leftrightarrow x=2\)

c)CM \(M⋮196\Leftrightarrow x=3\)

Answer 1

[Phùng Minh Quân]

\(a)\) \(M_{\left(3\right)}=3+3^2+3^3+...+3^{2016}\)

\(3M_{\left(3\right)}=3^2+3^3+3^4+...+3^{2017}\)

\(3M_{\left(3\right)}-M_{\left(3\right)}=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)

\(2M_{\left(3\right)}=3^{2017}-3\)

\(M_{\left(3\right)}=\frac{3^{2017}-3}{2}\)

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 ~ 

Answer 2

[Phùng Minh Quân]

\(b)\) 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 ~