Question

chung minh

\(\frac{a}{3}\)+\(\frac{a^2}{2}\)+\(\frac{a^3}{6}\) thuoc Z voi moi a thuoc Z

Answer 1

\(\frac{a}{3}+\frac{a^2}{2}+\frac{a^3}{6}=\frac{2a}{6}+\frac{3a^2}{6}+\frac{a^3}{6}=\frac{2a+3a^2+a^3}{6}\)

\(=\frac{a^3+3a^2+2a}{6}=\frac{a^3+2a^2+a^2+2a}{6}\)

\(=\frac{a^2.\left(a+2\right)+a.\left(a+2\right)}{6}=\frac{\left(a+2\right).\left(a^2+a\right)}{6}=\frac{\left(a+2\right).a.\left(a+1\right)}{6}\)

Vì a.(a+1).(a+2) là tích 3 số nguyên liên tiếp nên chia hết cho 2 và ;mà (2;3)=1

=>a.(a+1).(a+2) chia hết cho 6

\(=>\frac{a.\left(a+1\right).\left(a+2\right)}{6}\in Z\left(a\in Z\right)\) (đpcm)