题目内容
在(2x-3y)10的展开式中,求:(1)二项式系数的和;
(2)奇数项的二项式系数和与偶数项的二项式系数和;
(3)奇数项系数和与偶数项系数和.
【答案】分析:(1)由(1+1)10=
+…+
可求得结果;
(2)根据奇数项的二项式系数和等于偶数项的二项式系数和可得结果;
(3)设(2x-3y)10=
+
+
+…+
,分别令x=y=1;x=1,y=-1可得两式,联立即可求得答案;
解答:解:(1)(2x-3y)10的二项式系数和为
+…+
=(1+1)10=210;
(2)由于奇数项的二项式系数和等于偶数项的二项式系数和,
所以,奇数项的二项式系数和为
+
+…+
=29,
偶数项的二项式系数和为
+
+…
=29.
(3)设(2x-3y)10=
+
+
+…+
,
令x=y=1,得到a+a1+a2+…a10=1①,
令x=1,y=-1(或x=-1,y=1)得a-a1+a2-a3+…+a10=510②,
①+②得2(a+a2+…a10)=1+510,∴奇数项的系数和为
,
①-②得2(a1+a3+a9)=1-510,
∴偶数项的系数和为
.
点评:本题考查二项式定理的应用,考查学生的运算求解能力,属中档题.
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(2)根据奇数项的二项式系数和等于偶数项的二项式系数和可得结果;
(3)设(2x-3y)10=
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解答:解:(1)(2x-3y)10的二项式系数和为
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(2)由于奇数项的二项式系数和等于偶数项的二项式系数和,
所以,奇数项的二项式系数和为
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偶数项的二项式系数和为
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(3)设(2x-3y)10=
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令x=y=1,得到a+a1+a2+…a10=1①,
令x=1,y=-1(或x=-1,y=1)得a-a1+a2-a3+…+a10=510②,
①+②得2(a+a2+…a10)=1+510,∴奇数项的系数和为
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①-②得2(a1+a3+a9)=1-510,
∴偶数项的系数和为

点评:本题考查二项式定理的应用,考查学生的运算求解能力,属中档题.
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