观察下列各式:(a-1)(a+1)=a2-1(a-1)(a2+a+1)=a3+a2+a-a2-a-1=a3-1(a-1)(a3+a2+a+1)=a4+a3+a2+a-a3-a2-a-1=a4-1根据观

题目简介

观察下列各式:(a-1)(a+1)=a2-1(a-1)(a2+a+1)=a3+a2+a-a2-a-1=a3-1(a-1)(a3+a2+a+1)=a4+a3+a2+a-a3-a2-a-1=a4-1根据观

题目详情

观察下列各式:
(a-1)(a+1)=a2-1
(a-1)(a2+a+1)=a3+a2+a-a2-a-1=a3-1
(a-1)(a3+a2+a+1)=a4+a3+a2+a-a3-a2-a-1=a4-1
根据观察的规律,解答下列问题:
(1)填空:
①(a-1)(______)=a6-1;
②(a-1)(a11+a10+…+a+1)=______;
③(a-1)(an+an-1+an-2+…+a+1)=______.
(2)已知:1+22+24+26+…+22006+22008+22010=
1
3
×41006-
1
3

求:2+23+25+27+…+22007+22009的值.
题型:解答题难度:中档来源:不详

答案

(1)∵a-1)(a+1)=a2-1,
(a-1)(a2+a+1)=a3+a2+a-a2-a-1=a3-1,
(a-1)(a3+a2+a+1)=a4+a3+a2+a-a3-a2-a-1=a4-1,
∴①a5+a4+a3+a2+a+1;
②a12-1;
③an+1-1;
(2)因为(2-1)(1+2+22+23+24+…+22008+22009+22010)=22011-1,
即1+2+22+23+24+…+22008+22009+22010=22011-1.
1+22+24+26++22006+22008+22010=class="stub"1
3
×41006-class="stub"1
3

所以2+23+25+27++22007+22009=21011-1-(class="stub"1
3
×41006-class="stub"1
3
)

=22011-class="stub"1
3
×41006-class="stub"2
3
=class="stub"2
3
×41005-class="stub"2
3

故答案为:a5+a4+a3+a2+a+1,a12-1,an+1-1.

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