已知方程1.x²-3x+2=0两根之积为s1;2.x²-5x+6=0两根之积为s2;.n.x²-(2n+1)x+n(n+1)=0的两根之积为sn.求s1分之1+s2分之1+.+s2008分之1的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 21:59:55
![已知方程1.x²-3x+2=0两根之积为s1;2.x²-5x+6=0两根之积为s2;.n.x²-(2n+1)x+n(n+1)=0的两根之积为sn.求s1分之1+s2分之1+.+s2008分之1的值](/uploads/image/z/5442191-71-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8B1.x%26%23178%3B-3x%2B2%3D0%E4%B8%A4%E6%A0%B9%E4%B9%8B%E7%A7%AF%E4%B8%BAs1%EF%BC%9B2.x%26%23178%3B-5x%2B6%3D0%E4%B8%A4%E6%A0%B9%E4%B9%8B%E7%A7%AF%E4%B8%BAs2%EF%BC%9B.n.x%26%23178%3B-%EF%BC%882n%2B1%EF%BC%89x%2Bn%EF%BC%88n%2B1%EF%BC%89%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E4%B9%8B%E7%A7%AF%E4%B8%BAsn.%E6%B1%82s1%E5%88%86%E4%B9%8B1%2Bs2%E5%88%86%E4%B9%8B1%2B.%2Bs2008%E5%88%86%E4%B9%8B1%E7%9A%84%E5%80%BC)
已知方程1.x²-3x+2=0两根之积为s1;2.x²-5x+6=0两根之积为s2;.n.x²-(2n+1)x+n(n+1)=0的两根之积为sn.求s1分之1+s2分之1+.+s2008分之1的值
已知方程1.x²-3x+2=0两根之积为s1;2.x²-5x+6=0两根之积为s2;.n.x²-(2n+1)x+n(n+1)=0的两根之积为sn.求s1分之1+s2分之1+.+s2008分之1的值
已知方程1.x²-3x+2=0两根之积为s1;2.x²-5x+6=0两根之积为s2;.n.x²-(2n+1)x+n(n+1)=0的两根之积为sn.求s1分之1+s2分之1+.+s2008分之1的值
依题意知 s1=2,s2=6 .sn=n(n+1)
∴1/s1+1/s2+.+1/s2008=1/2+1/6+.1/2008*2009
=1/2+(1/2-1/3)+(1/3-1/4)+.+(1/2008-1/2009)
=1-1/2009
=2008/2009
s1=1x2,s2=2x3,s3=3x4.....s2008=2008x2009
s1分之1+s2分之1+......+s2008分之1
=1-1/2+1/2-1/3+1/3-1/4+........+1/2008-1/2009
=1-1/2009
=2008/2009
两根之积是sn=n(n+1)
所以1/s1+1/s2+...+1/s2008=1/1-1/2+1/2-1/3+....1/n-1/(n+1)=1-1/(n+1)=n/(n+1)=2008/2009
s1=1*2 s2=2*3 s3=3*4 ....... s2008=2008*2009
1/s1+1/s2+1/s3+........+1/s2008
=1/1*2+1/2*3+1/3*4+ ...... 1/2008*2009
=1-1/2+1/2-1/3+1/3-1/4+......+1/2008-1/2009
=1-1/2009
=2008/2009