= k(k+ 1) -k

=( 1/3)[k(k+ 1)(k+2)-(k- 1)k(k+ 1)]-( 1/2)[k(k+ 1)-(k- 1)k]

∑(k: 1->； k^2

=∑(k: 1-& gt； n){( 1/3)[k(k+ 1)(k+2)-(k- 1)k(k+ 1)]-( 1/2)[k(k+ 1)-(k- 1)k]}

=( 1/3)n(n+ 1)(n+2)-( 1/2)n(n+ 1)

=( 1/6)n(n+ 1)(2n+4-3)

=( 1/6)n(n+ 1)(2n+ 1)