(1+1/1+3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/98*100)*(1+1/99*101)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/05 09:19:55
(1+1/1+3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/98*100)*(1+1/99*101)
规律:1+1/1*3=1*3+1/1*3=4/1*3=2^2/1*3
1+1/3*4=2*4+1/2*4=9/2*4=3^2/2*4
1+1/3*5=3*5+1/3*5=16/3*5=4^2/3*5
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根据规律,可猜测:1+1/(2n-1)*(2n+1)=____________(n为整数)
1+1/2n*(2n+2)=__________(n为正整数)
根据规律计算:(1+1/1+3)*(1+1/2*4)*(1+13*5)*(1+1/4*6)*.*(1+1/18*100)*(1+1/99*101)
规律:1+1/1*3=1*3+1/1*3=4/1*3=2^2/1*3
1+1/3*4=2*4+1/2*4=9/2*4=3^2/2*4
1+1/3*5=3*5+1/3*5=16/3*5=4^2/3*5
.
根据规律,可猜测:1+1/(2n-1)*(2n+1)=____________(n为整数)
1+1/2n*(2n+2)=__________(n为正整数)
根据规律计算:(1+1/1+3)*(1+1/2*4)*(1+13*5)*(1+1/4*6)*.*(1+1/18*100)*(1+1/99*101)
1+1/(2n-1)*(2n+1)=(2n)^2/(2n-1)*(2n+1)(n为整数)
1+1/2n*(2n+2)=(2n+1)^2/2n*(2n+2)(n为正整数)
(1+1/1*3)*(1+1/2*4)*(1+/3*5)*(1+1/4*6)*.*(1+1/98*100)*(1+1/99*101)=200/101
1+1/2n*(2n+2)=(2n+1)^2/2n*(2n+2)(n为正整数)
(1+1/1*3)*(1+1/2*4)*(1+/3*5)*(1+1/4*6)*.*(1+1/98*100)*(1+1/99*101)=200/101
(1+1/1*3)*(1+1/2*4)*(1+1/3*5).(1+1/98*100)*(1+1/99*101)
(1+1/1+3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/98*100)*(1+1/99*101)
(1+1*3分之1)(1+2*4分之1)(1+3*5分之1)···(1+98*100分之1)(1+99*101分之1)
计算;(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/98*100)*(1+1/99*101)
1/1*2*3+1/2*3*4+1/3*4*5+...+1/99*100*101
1/1*2*3+1/2*3*4+1/3*4*5+.1/99*100*101
计算:101-100+99-98+...+5-4+3-2+1
1/1×2+1/2×3+1/3×4+1/4×5+.+1/98×99+1/99×100=?
1/1*2+1/2*3+1/3*4+1/4*5+.1/98*99+1/99*100=?
1*2/1+2*3/1+3*4/1+.100*101/1 1*3/1+3*5/1+5*7/1+.+99*101/1 怎么
要简算 1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100+1/99*100*101
1*2/1+2*3/1+3*4/1+.98*99/1+99*100/1=?