Chọn D.

Với ∀x ∈ R ta có: 

Lấy đạo hàm hai vế theo x ta được:

Thay x = -2 vào (1) ta được:

Từ yêu cầu bài toán ta có: 2n + 1 2017 ⇔ n = 2018.

Đáp án A

y = x + x 2 + x 3 + ... + x 2018 = x 1 − x 2018 1 − x y ' = 1 + 2 x + 3 x 2 + ... + 2018 x 2017         = 2018 x 2019 − 2019 x 2018 + 1 ( 1 − x ) 2 y ' ( 2 ) = 1 + 2.2 + 3.2 2 + ... + 2018.2 2017 = 2017.2 2018 + 1

Đáp án là A

Ta có: \(\frac{1}{1\cdot2001}+\frac{1}{2\cdot2002}+\cdots+\frac{1}{19\cdot2019}\)

\(=\frac{1}{2000}\left(\frac{2000}{1\cdot2001}+\frac{2000}{2\cdot2002}+\cdots+\frac{2000}{19\cdot2019}\right)\)

\(=\frac{1}{2000}\left(1+\frac12+\cdots+\frac{1}{19}-\frac{1}{2001}-\frac{1}{2002}-\cdots-\frac{1}{2019}\right)\)

Ta có: \(\frac{1}{1\cdot20}+\frac{1}{2\cdot21}+\cdots+\frac{1}{2000\cdot2019}\)

\(=\frac{1}{19}\left(\frac{19}{1\cdot20}+\frac{19}{2\cdot21}+...+\frac{19}{2000\cdot2019}\right)\)

\(=\frac{1}{19}\left(1-\frac{1}{20}+\frac12-\frac{1}{21}+\cdots+\frac{1}{2000}-\frac{1}{2019}\right)\)

\(=\frac{1}{19}\left(1+\frac12+\cdots+\frac{1}{19}-\frac{1}{2001}-\frac{1}{2002}-\cdots-\frac{1}{2019}\right)\)

TA có: \(B=\frac{\frac{1}{1\cdot2001}+\frac{1}{2\cdot2002}+\cdots+\frac{1}{19\cdot2019}}{\frac{1}{1\cdot20}+\frac{1}{2\cdot21}+\cdots+\frac{1}{2000\cdot2019}}\)

\(=\frac{1}{2000}:\frac{1}{19}=\frac{19}{2000}\)

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{99}\right)⋮3\)

\(M=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)

Đặt  \(t=a^2+5a+5\)

\(M=\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2=\left(a^2+5a+5\right)^2\)