2) k=0 代入(1)得 ∑k(k-)lcx2-∑k(-1)cx2-2∑kcx2 k=1 1(7+1)∑cx=0 x0:2·l2+(1+1)cn=0 (+1) 2·1 00 2 k k k y c x ¥ = = å （） 代入(1),得 ( ) ( ) -2 2 2 1 -1 - -1 - 2 k k k k k k k k k k k c x k k c x kc x ¥ ¥ ¥ = = = å å å ( ) 0 1 0 k k k l l c x ¥ = + + = å 0 2 0 x : 2 × lc +（l l c + = 1) 0 2 0 ( 1) 2 1 l l c c + \ = ×