a) A < B              

b) C > D

VÌ 20192019+120192020 +1=140384040 >20192018+120192019 =140384038 nên A>B

N > M nhé bạn.

Đáp án B

Đặt t = log u 1 , khi đó giả thiết ⇔ t 3 - 2 t 2 + t - 2 = 0 ⇔ t - 2 t 2 + 1 = 0 ⇔ t = 2 ⇒ log u 1 = 2  

Ta có u n + 1 = 2 u n + 10 ⇔ u n + 1 + 10 = 2 u n + 10 ⇔ v n + 1 = 2 v n  với v n = u n + 10  

Dễ thấy v n + 1 = 2 v n  là một cấp số nhân với công bội q = 2 ⇒ v n = v 1 . 2 n - 1  

Mà log u 1 = 2 ⇒ u 1 = 10 2 = 100  suy ra v 1 = u 1 + 10 = 110 ⇒ v n = 100 . 2 n - 1  

Khi đó u n = v n - 10 = 100 . 2 n - 1 - 10 > 10 100 - 10 ⇔ 2 n - 1 > 10 98 ⇔ n > log 2 10 98 + 1 = 326 , 54  

Vậy giá trị nhỏ nhất của n cần tìm là n m i n = 327 .

d) Đ

\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{1008}+1\right)=2^{2016}-1< 2^{2016}=M\)

\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)

\(N=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ N=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=M\)

Ta có: \(M=1\cdot2014+2\cdot2013+\cdots+2014\cdot1\)

\(=2\left(1\times2014+2\times2013+\cdots+1007\times1008\right)\)

\(=2\left\lbrack1\times\left(2015-1\right)+2\times\left(2015-2\right)+\cdots+1007\times\left(2015-1007\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times\left(1+2+\cdots+1007\right)-\left(1^2+2^2+\cdots+1007^2\right)\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times\frac{1008}{2}-\frac{1007\times\left(1007+1\right)\times\left(2\times1007+1\right)}{6}\right\rbrack\)

\(=2\cdot\left\lbrack2015\times1007\times504-1007\times168\times2015\right\rbrack=2\times2015\times1007\times168\left(3-1\right)=4\times168\times2015\times1007\)

\(N=1+\left(1+2\right)+\cdots+\left(1+2+\cdots+2014\right)\)

\(\) \(=\frac{1\times2}{2}+\frac{2\times3}{2}+\cdots+\frac{2014\times2015}{2}\)

\(=\frac12\times\left(1\times2+2\times3+\cdots+2014\times2015\right)\)

\(=\frac12\times\left\lbrack1\times\left(1+1\right)+2\times\left(2+1\right)+\cdots+2014\times\left(2014+1\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\left(1\times1+2\times2+\cdots+2014\times2014\right)+\left(1+2+\cdots+2014\right)\right\rbrack\)

\(=\frac12\times\left\lbrack\frac{2014\times\left(2014+1\right)\times\left(2\times2014+1\right)}{6}+\frac{2014\times2015}{2}\right\rbrack\)

\(=\frac12\times\left\lbrack1007\times2015\times1343+1007\times2015\right\rbrack=\frac12\times1007\times2015\times\left(1343+1\right)\)

=1007x2015x672

=4x168x2015x1007

Do đó: M=N