由题意可得,,由bn+cn=1可得bn+1=bn+1(bn+cn)=bn+1bn+bn+1Cn=bn+1bn+cn+1=bnbn+1+1-bn+1即2bn+1-bnbn+1-1=0,则bn+1-1=bn+1(bn-1)=(bn-1)(bn+1-1+1)=(bn-1)(bn+1-1)+(bn-1),从而可得 ,由等差数列的通项公式可得,可求 ,利用递推公式an=bn-bn-1可求an
【解析】
由题意可得,
bn+cn=1
∴bn+1=bn+1(bn+cn)=bn+1bn+bn+1Cn
=bn+1bn+cn+1=bnbn+1+1-bn+1
∴2bn+1-bnbn+1-1=0
∴bn+1(2-bn)=1
∴0<bn<2
若bn+1=1则bn=1,bn-1=bn-2=…=b1=1与 矛盾
∴bn+1≠1
∴bn+1-1=bn+1(bn-1)
=(bn-1)(bn+1-1+1)
=(bn-1)(bn+1-1)+(bn-1)
∴
∴且
∴是以-2为首项,以-1为公差的等差数列
由等差数列的通项公式可得,=-n-1
∴
∴an=bn-bn-1==
∴
所以的前10项之和等于12+22+…+102+(1+2+3+…+10)=440
故答案为:440.