摘要:设是方程①的两个不同的根.
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设代数方程a0-a1x2+a2x4-…+(-1)nanx2n=0有2n个不同的根±x1,±x2,…,±xn,则a0-a1x2+a2x4-…+(-1)nanx2n=a0(1-
)(1-
)•…•(1-
),比较两边x2的系数得a1=
=1-
+
-
+…对x∈R,x≠0成立,则由于
=0有无穷多个根:±π,±2π,…,+±nπ,…,于是1-
+
-
+…=(1-
)(1-
)•…•(1-
)•…,利用上述结论可得1+
+
+…+
+…=
.
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| x2 | ||
|
| x2 | ||
|
| x2 | ||
|
a0(
+
+…+
)
| 1 | ||
|
| 1 | ||
|
| 1 | ||
|
a0(
+
+…+
)
(用a0•x1•x2•…•xn表示);若已知展开式| 1 | ||
|
| 1 | ||
|
| 1 | ||
|
| sinx |
| x |
| x2 |
| 3! |
| x4 |
| 5! |
| x6 |
| 7! |
| sinx |
| x |
| x2 |
| 3! |
| x4 |
| 5! |
| x6 |
| 7! |
| x2 |
| π2 |
| x2 |
| 22•π2 |
| x2 |
| n2π2 |
| 1 |
| 22 |
| 1 |
| 32 |
| 1 |
| n2 |
| π2 |
| 6 |
| π2 |
| 6 |