\(\)Đáp án: \(\dfrac{2013.2014.2015}{3}\)
Tổng quát: \(S_n=1.2+2.3+...\left(n-1\right).n\)
Ta sẽ chứng minh \(S_n=\dfrac{\left(n-1\right)n\left(n+1\right)}{3}\) với mọi n nguyên, \(n\ge2\) bằng quy nạp.
- Với \(n=2:S_2=1.2=2=\dfrac{1.2.3}{3}\)
- Giả sử khẳng định đúng đến \(n=k:S_k=\dfrac{\left(k-1\right)k\left(k+1\right)}{3}\)
- Với \(n=k+1:\)
\(S_{k+1}=1.2+2.3+...+\left(k-1\right).k+k.\left(k+1\right)\\ =S_k+k.\left(k+1\right)\\ =\dfrac{\left(k-1\right).k.\left(k+1\right)}{3}+k.\left(k+1\right)\\ =\dfrac{\left(k-1\right).k.\left(k+1\right)+3.k.\left(k+1\right)}{3}\\ =\dfrac{k.\left(k+1\right).\left(k+2\right)}{3}\left(\text{dpcm}\right)\)
Vậy \(D=S_{2014}=\dfrac{2013.2014.2015}{3}\)
