Question

cho n chan. chứng minh\(\dfrac{n^3}{24}\)+\(\dfrac{n^2}{8}\)+\(\dfrac{n}{12}\)∈ Z

Answer 1

\(=\dfrac{n^3+3n^2+2n}{24}=\dfrac{n\left(n+1\right)\left(n+2\right)}{24}\)

\(=\dfrac{2k\left(2k+1\right)\left(2k+2\right)}{24}=\dfrac{4k\left(2k+1\right)\left(k+1\right)}{24}\)

\(=\dfrac{4k\left(k+1\right)\left(k+2+k-1\right)}{24}\)

\(=\dfrac{4k\left(k+1\right)\left(k+2\right)+4k\left(k+1\right)\left(k-1\right)}{24}=\dfrac{k\left(k+1\right)\left(k+2\right)+k\left(k+1\right)\left(k-1\right)}{6}\)

Vì k;k+1;k+2 là ba số liên tiếp

nen k(k+1)(k+2) chia hết cho 3!=6

k;k+1;k-1 là ba số liên tiếp

nên k(k+1)(k-1) chia hết cho 3!=6

=>A chia hêt cho 6