题目内容
(理科做) 用数学归纳法证明:
+
+…+
=
.
| 12 |
| 1•3 |
| 22 |
| 3•5 |
| n2 |
| (2n-1)(2n+1) |
| n(n+1) |
| 2(2n+1) |
证明:(1)当n=1时,左=
=右,等式成立.
(2)假设当n=k时等式成立,
即
+
+…+
=
,
当n=k+1时,左边=
+
+…+
+
=
+
=
.
∴当n=k+1时,等式也成立.
综合(1)(2),等式对所有正整数都成立.
| 1 |
| 3 |
(2)假设当n=k时等式成立,
即
| 12 |
| 1•3 |
| 22 |
| 3•5 |
| k2 |
| (2k-1)(2k+1) |
| k(k+1) |
| 2(2k+1) |
当n=k+1时,左边=
| 12 |
| 1•3 |
| 22 |
| 3•5 |
| k2 |
| (2k-1)(2k+1) |
| (k+1)2 |
| (2k+1)(2k+3) |
| k(k+1) |
| 2(2k+1) |
| (k+1)2 |
| (2k+1)(2k+3) |
| (k+1)(k+2) |
| 2(2k+3) |
∴当n=k+1时,等式也成立.
综合(1)(2),等式对所有正整数都成立.
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