根据杨辉三角,请计算(x-1/x)的12次方的展开式中四次项的系数是? (x+1+1/x)^4展开式中含X^2项的系数是?[有追加分...
\uff08x+1/x-2\uff09\u76844\u6b21\u65b9\u7684\u5c55\u5f00\u5f0f\u4e2dx\u7684\u5e73\u65b9\u7684\u7cfb\u6570\uff1f\u6839\u636e\u4e8c\u9879\u5f0f\u5b9a\u7406\uff0c\u542b x^2 \u9879\u7684\u7cfb\u6570\u4e3a C(5\uff0c2) (-2)^3 = - 80 \u3002
\u8fd9\u4e2a\u9898\u76ee\u662f\u521d\u4e2d\u9898\u76ee\u7684\u8bdd\uff0c\u90a3\u4e48\u628a
x
+
1/x
\u7ec4\u5408\u5728\u4e00\u8d77\uff0c\u76f4\u63a5\u5c55\u5f00\u3002
(a+b)^n
\u7684\u5c55\u5f00\u7cfb\u6570
\u53ef\u7528
\u201c
\u6768\u8f89\u4e09\u89d2\u201d\uff0c\u5982\u4e0b\u56fe\uff1a
1
1
1
2
1
1
3
3
1
1
4
6
4
1
\u4f59\u6b64\u7c7b\u63a8
\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d\uff0d
(x
+
1
+
1/x)^4
=
[(x
+
1/x)
+
1]^4
=
(x
+
1/x)^4
+
4(x
+
1/x)^3
+
6(x
+
1/x)^2
+
4
(x
+
1/x)
+
1
(x
+
1/x)^4
\u5c55\u5f00\u5f0f\u7b2c2\u9879\u662f
x^3
*
1/x\uff0c\u5373
x^2\u9879\uff0c\u7cfb\u6570\u4e3a
4
(x
+
1/x)^3
\u5c55\u5f00\u5f0f\u4e2d\u65e0
x^2
\u9879
(x
+
1/x)^2
\u5c55\u5f00\u5f0f\u4e2d\uff0c\u7b2c\u4e00\u9879\u4e3a
x^2
\u9879\uff0c\u7cfb\u6570\u4e3a
1
\u56e0\u6b64
(x+1+1/x)^4\u5c55\u5f00\u5f0f\u4e2d\u542bX^2\u9879\u7684\u7cfb\u6570\u662f
4
+
6\u00d71
=
10
=495x^4,
所以系数是 495。
绛旓細鍥涙椤规槸 C(12锛8)x^8(-1/x)^4 锛495x^4锛屾墍浠ョ郴鏁版槸 495銆
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