Ta có: \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\)
\(\Rightarrow1^3-1+2^3-2+...+50^3-50\)
\(=0+1.2.3+2.3.4+...+49.50.51\)
\(=\frac{49.50.51.52}{4}=1624350\)
Ta lại có:
\(1+2+3+...+50=\frac{50.51}{2}=1275\)
\(\Rightarrow1^3+2^3+...+50^3=1624350+1275=1625625=1275^2\)
Vậy nó chia hết cho 1275
Nhận xét : \(k^3=\left[\frac{k\left(k+1\right)}{2}\right]^2-\left[\frac{k\left(k-1\right)}{2}\right]^2\)
Tương tự,thế vào ta có :
\(1^3+2^3+...+50^3=-\left(\frac{1\cdot2}{2}\right)^2+\left(\frac{1\cdot0}{2}\right)^2-\left(\frac{2\cdot3}{2}\right)^2+\left(\frac{2\cdot1}{2}\right)^2-...\)
\(-\left(\frac{50\cdot51}{2}\right)^2+\left(\frac{50\cdot49}{2}\right)^2\)
\(=\left[\frac{50\left(50-1\right)}{2}\right]^2\)
\(=\left(1+2+3+...+50\right)^2⋮\left(1+2+3+..+50\right)\)
Mà \(1+2+3+...+50=1275\)
=> Ta có đpcm
