解题思路:由题意可知Cn=An×Bn-An-1×Bn-1,Cn=C1+C2+…+Cn-1+Cn=a1×b1+(A2×B2-a1×b1)+…+(An-1×Bn-1-An-2×Bn-2)+(An×Bn-An-1×Bn-1)=An×Bn,由此可以求出数列{Cn}的前100项的和.
Cn=an•Bn+bn•An-an•bn
=(An-An-1)×Bn+(Bn-Bn-1)×An-(An-An-1)×(Bn-Bn-1)
=An×Bn-An-1×Bn+Bn×An-Bn-1×An-(An×Bn-An-1×Bn-An×Bn-1+An-1×Bn-1]
=An×Bn-An-1×Bn-1,
∴Cn=An×Bn-An-1×Bn-1,
Cn-1=An-1×Bn-1-An-2×Bn-2
…C2=A2×B2-a1×b1
C1=a1×b1
∴Cn=C1+C2+…+Cn-1+Cn
=a1×b1+(A2×B2-a1×b1)+…+(An-1×Bn-1-An-2×Bn-2)+(An×Bn-An-1×Bn-1)=An×Bn
∴C100=A100×B100=8×251=2008
C(100)=A(100)×B(100)=8×251=2008.
答案:2008.
点评:
本题考点: 数列的应用.
考点点评: 本题考查数列的性质和应用,解题时要注意培养学生的计算能力.