>>393
a(k+1)−a(k)=k^2。
 a(n)
=a(1)+Σ_{1≦k≦n−1}(a(k+1)−a(k))
=1+Σ_{1≦k≦n−1}(k^2)
=1+(n−1)((n−1)+1)(2(n−1)+1)/6
=1+n(n−1)(2n−1)/6。