优质解答
(1)依题可知B2B3=6i+6j由OB1⊥B2B3知6a-36=0,所以a=6;…(4分)(2)OBn=OB1+B1B2+…+Bn−1Bn…(2分)=(a,-6)+(6,3)+(6,3•2)+…+(6,3•2n-2)=(6n+a-6,3•2n-1-9)所以OBn=(6n+a−6,3•2n−1...
(1)依题可知B2B3=6i+6j由OB1⊥B2B3知6a-36=0,所以a=6;…(4分)(2)OBn=OB1+B1B2+…+Bn−1Bn…(2分)=(a,-6)+(6,3)+(6,3•2)+…+(6,3•2n-2)=(6n+a-6,3•2n-1-9)所以OBn=(6n+a−6,3•2n−1...