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已知数列{an},{bn},它们的前n项和分别为An,Bn,记cn=anBn+bnAn-anbn(n∈N*),则数列{cn}的前10项和为()A.A10+B10B.12(A10+B10)C.A10•B10D.A10•B10
题目详情
已知数列{an},{bn},它们的前n项和分别为An,Bn,记cn=anBn+bnAn-anbn(n∈N*),则数列{cn}的前10项和为( )
A.A10+B10
B.
(A10+B10)
C.A10•B10
D.
A.A10+B10
B.
1 |
2 |
C.A10•B10
D.
A10•B10 |
▼优质解答
答案和解析
∵an=An-An-1,bn=Bn-Bn-1,
cn=anBn+bnAn-anbn(n∈N*),
∴cn=an(Bn-bn)+bnAn
=(An-An-1)(Bn-bn)+bnAn
=AnBn-An-1(Bn-bn)
=AnBn-An-1(Bn-Bn+Bn-1)
=AnBn-An-1Bn-1,
∴数列{cn}的前10项和为:
c1+c2+c3+…+c10
=A1B1+(A2B2-A1B1)+(A3B3-A2B2)+…+(A10B10-A9B9)
=A10B10.
故选:C.
cn=anBn+bnAn-anbn(n∈N*),
∴cn=an(Bn-bn)+bnAn
=(An-An-1)(Bn-bn)+bnAn
=AnBn-An-1(Bn-bn)
=AnBn-An-1(Bn-Bn+Bn-1)
=AnBn-An-1Bn-1,
∴数列{cn}的前10项和为:
c1+c2+c3+…+c10
=A1B1+(A2B2-A1B1)+(A3B3-A2B2)+…+(A10B10-A9B9)
=A10B10.
故选:C.
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