Question

求和S=1×2×3+2×3×4+…+n(n+1)(n+2).

Answer 1

解法一：研究通項(xiàng).

a_k=k(k+1)(k+2)=k³+3k²+2k.

∴S=(1³+2³+…+n³)+3(1²+2²+…+n²)+2(1+2+3+…+n)

=[]²+3×+2×=n(n+1)(n+2)(n+3).

解法二：通項(xiàng)a_k=6.

∴S=6(++…++2)

=6+3=n(n+1)(n+2)(n+3).