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(1)填空:(a-b)(a+b)=;(a-b)(a2+ab+b2)=;(a-b)(a3+a2b+ab2+b3)=;(2)猜想:(a-b)(an-1+an-2b+an-3b2+…+abn-2+bn-1)=(其中n为正整数,且n≥2);(3)利用(2)猜想的结论计
题目详情
(1)填空:
(a-b)(a+b)=___;
(a-b)(a2+ab+b2)=___;
(a-b)(a3+a2b+ab2+b3)=___;
(2)猜想:
(a-b)(an-1+an-2b+an-3b2+…+abn-2+bn-1)=___(其中n为正整数,且n≥2);
(3)利用(2)猜想的结论计算:①29+28+27+…+22+2+1
②210-29+28-…-23+22-2.
(a-b)(a+b)=___;
(a-b)(a2+ab+b2)=___;
(a-b)(a3+a2b+ab2+b3)=___;
(2)猜想:
(a-b)(an-1+an-2b+an-3b2+…+abn-2+bn-1)=___(其中n为正整数,且n≥2);
(3)利用(2)猜想的结论计算:①29+28+27+…+22+2+1
②210-29+28-…-23+22-2.
▼优质解答
答案和解析
(1)(a-b)(a+b)=a2-b2;
(a-b)(a2+ab+b2)=a3+a2b+ab2-a2b-ab2-b3=a3-b3;
(a-b)(a3+a2b+ab2+b3)=a4+a3b+a2b2+ab3-a3b-a2b2-ab3-b4=a4-b4;
故答案为:a2-b2,a3-b3,a4-b4;
(2)由(1)的规律可得:原式=an-bn,
故答案为:an-bn;
(3)①29+28+27+…+23+22+2+1=(2-1)•(29+28•1+27•12+…+23•16+22•17+2•18)+1
=210-110+1
=210-1+1
=1024;
②210-29+28-…+23-22+2
=
×[2-(-1)]•[210×(-1)0+29×(-1)1+27×(-1)2+…+23×(-1)7+22×(-1)8+2×(-1)9+20×(-1)10-1]
=
[211-(-1)11-1]
=
×2048
=
.
(a-b)(a2+ab+b2)=a3+a2b+ab2-a2b-ab2-b3=a3-b3;
(a-b)(a3+a2b+ab2+b3)=a4+a3b+a2b2+ab3-a3b-a2b2-ab3-b4=a4-b4;
故答案为:a2-b2,a3-b3,a4-b4;
(2)由(1)的规律可得:原式=an-bn,
故答案为:an-bn;
(3)①29+28+27+…+23+22+2+1=(2-1)•(29+28•1+27•12+…+23•16+22•17+2•18)+1
=210-110+1
=210-1+1
=1024;
②210-29+28-…+23-22+2
=
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