第二,三题请告诉我简便计算
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/08/14 15:11:21
第二,三题请告诉我简便计算
![](http://img.wesiedu.com/upload/e/32/e3247b689e15741f79911abcf1bf9566.jpg)
![](http://img.wesiedu.com/upload/e/32/e3247b689e15741f79911abcf1bf9566.jpg)
![第二,三题请告诉我简便计算](/uploads/image/z/17472221-53-1.jpg?t=%E7%AC%AC%E4%BA%8C%2C%E4%B8%89%E9%A2%98%E8%AF%B7%E5%91%8A%E8%AF%89%E6%88%91%E7%AE%80%E4%BE%BF%E8%AE%A1%E7%AE%97)
第二题解答如下:
1+2+3+...+n=n(n+1)/2
1/(1+2+3+...+n)=2/n(n+1)=2[1/n-1/(n+1)]
活用公式是关键.
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/(1+2+3+...+100)
=2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/100-1/101)]
=2(1-1/101)
=200/101
第三题如下:
关键是把又变为分数形式,提前公因子.
(1又1/2001 ×8.12+5又31/75÷2/3×8又2000/2001 )÷4又11/48
=(1又1/2001 ×203/25+406/75×3/2×8又2000/2001 )÷203/48
=(1又1/2001 ×203/25+203/25×8又2000/2001 )÷203/48
=203/25×(1又1/2001 +8又2000/2001 )÷203/48
=203/25×10÷203/48
=10/25×48
=96/5
1+2+3+...+n=n(n+1)/2
1/(1+2+3+...+n)=2/n(n+1)=2[1/n-1/(n+1)]
活用公式是关键.
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+.+1/(1+2+3+...+100)
=2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/100-1/101)]
=2(1-1/101)
=200/101
第三题如下:
关键是把又变为分数形式,提前公因子.
(1又1/2001 ×8.12+5又31/75÷2/3×8又2000/2001 )÷4又11/48
=(1又1/2001 ×203/25+406/75×3/2×8又2000/2001 )÷203/48
=(1又1/2001 ×203/25+203/25×8又2000/2001 )÷203/48
=203/25×(1又1/2001 +8又2000/2001 )÷203/48
=203/25×10÷203/48
=10/25×48
=96/5