Vì :

2/1 > 14 => A > 14

20/199>20 => A>20

https://lazi.vn/edu/exercise/289338/cho-b-2-1-x-4-3-x-6-5-x-8-x-x-200-199-chung-minh-rang-14-lt-b-lt-20

Em xem tại link này nhé (Chị gửi cho)

Học tốt

!!!!!!!!!

\(A=1\left(2+2\right)+2\left(2+3\right)+3\left(2+4\right)+.....+\left(n-1\right)\left(2+n\right)\)

\(\Leftrightarrow A=1.2+1.2+2.3+2.2+3.4+2.3+....+\left(n-1\right)n+2\left(n-1\right)\)

\(\Leftrightarrow A=\left(1.2+2.3+.....+\left(n-1\right)n\right)+2\left(1+2+3+....+\left(n-1\right)\right)\)

Giả sử A=B+C

Với \(\begin{cases}B=1.2+2.3+.....+\left(n-1\right)n\\C=2\left[1+2+....+\left(n-1\right)\right]\end{cases}\)

Ta có

\(3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+\left(n-1\right)n\left[\left(n+1\right)-\left(n-2\right)\right]\)

\(\Rightarrow3B=1.2.3-0.1.2+2.3.4-1.2.3+.....+\left(n-1\right)n\left(n+1\right)-\left(n-2\right)\left(n-1\right)n\)

\(\Rightarrow B=\frac{\left(n-1\right)n\left(n+1\right)}{3}\)

Mặt khác

\(C=2\left[1+2+....+\left(n-1\right)\right]\)

\(\Rightarrow C=2.\frac{\left[\left(n-1\right)+1\right]n}{2}=n^2\)

\(\Rightarrow A=\frac{\left(n-1\right)n\left(n+1\right)}{3}+n^2\)

Vậy \(A=\frac{\left(n-1\right)n\left(n+1\right)}{3}+n^2\)

\(A=1\cdot4+2\cdot5+3\cdot6+...+n\left(n+3\right)\)

\(=1\left(1+3\right)+2\left(2+3\right)+3\left(3+3\right)+...+n\left(n+3\right)\)

\(=\left(1^2+2^2+...+n^2\right)+3\left(1+2+3+...+n\right)\)

\(=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+3\cdot\dfrac{n\left(n+1\right)}{2}\)

\(=\dfrac{n\left(n+1\right)\left(2n+1\right)+9n\left(n+1\right)}{6}\)

\(=\dfrac{n\left(n+1\right)\left(2n+1+9\right)}{6}\)

\(=\dfrac{n\left(n+1\right)\left(2n+10\right)}{6}=\dfrac{n\left(n+1\right)\left(n+5\right)}{3}\)

\(\frac{n\left(n+1\right)\left(n+5\right)}{3}\)