(1)分解因式:(x+1)(x+2)(x+3)-x(x+1)(x+2) (2)计算:1×2+2×3+3×4+.+n(n+
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/02 19:13:53
(1)分解因式:(x+1)(x+2)(x+3)-x(x+1)(x+2) (2)计算:1×2+2×3+3×4+.+n(n+1)(n为正整数)
(3)1*2*3+2*3*4+4*5*6+ …+24*25*26,并说明算法
(3)1*2*3+2*3*4+4*5*6+ …+24*25*26,并说明算法
(1)分解因式:(x+1)(x+2)(x+3)-x(x+1)(x+2)
(x+1)(x+2)(x+3)-x(x+1)(x+2)
=(x+1)(x+2)[(x+3)-x]
=3(x+1)(x+2)
(2)计算:1×2+2×3+3×4+……+n(n+1)(n为正整数)
由于n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
所以1*2+2*3+……+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+…….+n(n+1)(n+2)-(n-1)n(n+1)]/3
[前后消项]
=[n(n+1)(n+2)]/3
(3)1*2*3+2*3*4+3*4*5+……+24*25*26,并说明算法
因为4(n-1)n(n+1)=(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1)
所以
1*2*3+2*3*4+3*4*5+……+n(n+1)(n+2)
=n(n+1)(n+2)(n+3)/4
1*2*3+2*3*4+3*4*5+.+24*25*26
=24*25*26*27/4
=105300
(x+1)(x+2)(x+3)-x(x+1)(x+2)
=(x+1)(x+2)[(x+3)-x]
=3(x+1)(x+2)
(2)计算:1×2+2×3+3×4+……+n(n+1)(n为正整数)
由于n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
所以1*2+2*3+……+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+…….+n(n+1)(n+2)-(n-1)n(n+1)]/3
[前后消项]
=[n(n+1)(n+2)]/3
(3)1*2*3+2*3*4+3*4*5+……+24*25*26,并说明算法
因为4(n-1)n(n+1)=(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1)
所以
1*2*3+2*3*4+3*4*5+……+n(n+1)(n+2)
=n(n+1)(n+2)(n+3)/4
1*2*3+2*3*4+3*4*5+.+24*25*26
=24*25*26*27/4
=105300
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