题目内容
用数学归纳法证明:
+
+…+
=
(n∈N*).
| 12 |
| 1•3 |
| 22 |
| 3•5 |
| n2 |
| (2n-1)(2n+1) |
| n(n+1) |
| 2(2n+1) |
证明(1)n=1时,
左边
=
=
=右边,等式成立
(2)假设n=k时等式成立,
即
+
++
=
.
则n=k+1时,
左边=
+
=
(k+
)
=
•
=
•
=
.
∴n=k+1时,等式成立
由(1)(2)知,对一切n∈N*,
+
++
=
.
左边
| 12 |
| (2×1-1)(2×1+1) |
| 1 |
| 3 |
| 1×(1+1) |
| 2(2×1+1) |
(2)假设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) |
则n=k+1时,
左边=
| k(k+1) |
| 2(2k+1) |
| (k+1)2 |
| (2k+1)(2k+3) |
| k-1 |
| 2(2k+1) |
| 2k+2 |
| 2k+3 |
=
| k+1 |
| 2(2k+1) |
| 2k2+5k+2 |
| 2k+3 |
| k+1 |
| 2(2k+1) |
| (2k+1)(k+2) |
| 2k+3 |
| (k+1)(k+2) |
| 2(2k+3) |
∴n=k+1时,等式成立
由(1)(2)知,对一切n∈N*,
| 12 |
| 1•3 |
| 22 |
| 3•5 |
| n2 |
| (2n-1)(2n+1) |
| n(n+1) |
| 2(2n+1) |
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