Question

Tính D = 1.2 + 2.3 + 3.4 +...+ 2013.2014

Cần gấp

Answer 1

\(\)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 \(D=S_{2014}=\dfrac{2013.2014.2015}{3}\)