问题
解答题
用数学归纳法证明:
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答案
证明:(1)当n=1时,等式左边=
=1 2×4
,等式右边=1 8
=1 4(1+1)
,∴等式成立.1 8
(2)假设n=k(k≥1.k∈N*)时等式成立,
即
+1 2×4
+1 4×6
++1 6×8
=1 2k(2k+2)
成立,k 4(k+1)
那么当n=k+1时,
+1 2×4
+1 4×6
++1 6×8
+1 2k(2k+2) 1 2(k+1)[2(k+1)+2]
=
+k 4(k+1) 1 4(k+1)(k+2)
=k(k+2)+1 4(k+1)(k+2)
=(k+1)2 4(k+1)(k+2)
=k+1 4[(k+1)+1]
即n=k+1时等式成立.由(1)、(2)可知,对任意n∈N*等式均成立.