已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:1/(A3+1)(B3+1)+1/(A4+1)(B4+1)+……+1/(A20+1)(B20+1)
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![已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:1/(A3+1)(B3+1)+1/(A4+1)(B4+1)+……+1/(A20+1)(B20+1)](/uploads/image/z/6908540-68-0.jpg?t=%E5%B7%B2%E7%9F%A5n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E5%85%B3%E4%BA%8Ex%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bx%5E2%2B%282n%2B1%29%2Bn%5E2%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E4%B8%BAAn%E3%80%81Bn%2C%E6%B1%82%E4%B8%8B%E5%BC%8F%E7%9A%84%E5%80%BC%3A%E5%B7%B2%E7%9F%A5n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E5%85%B3%E4%BA%8Ex%E7%9A%84%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bx%5E2%2B%282n%2B1%29%2Bn%5E2%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E4%B8%BAAn%E3%80%81Bn%2C%E6%B1%82%E4%B8%8B%E5%BC%8F%E7%9A%84%E5%80%BC%EF%BC%9A1%2F%28A3%2B1%29%28B3%2B1%29%2B1%2F%28A4%2B1%29%28B4%2B1%29%2B%E2%80%A6%E2%80%A6%2B1%2F%28A20%2B1%29%28B20%2B1%29)
已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:1/(A3+1)(B3+1)+1/(A4+1)(B4+1)+……+1/(A20+1)(B20+1)
已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:
已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,
求下式的值:
1/(A3+1)(B3+1)+1/(A4+1)(B4+1)+……+1/(A20+1)(B20+1)
已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:已知n为正整数,关于x的二次方程x^2+(2n+1)+n^2=0的两根为An、Bn,求下式的值:1/(A3+1)(B3+1)+1/(A4+1)(B4+1)+……+1/(A20+1)(B20+1)
由二次方程根与系数的关系(也就是韦达定理)可得
An+Bn=-(2n+1) ,An*Bn=n^2 ,
因此 1/[(An+1)(Bn+1)]=1/(An*Bn+An+Bn+1)=1/(n^2-2n-1+1)=1/[n(n-2)]=1/2*[1/(n-2)-1/n] ,
所以,原式=1/(1*3)+1/(2*4)+.+1/(18*20)
=1/2*(1-1/3+1/2-1/4+1/3-1/5+.+1/18-1/20)
=1/2*(1+1/2-1/19-1/20)
=531/760 .
X2+1=-(n2+2n), An*Bn=X2, An+Bn=0;
1/(An+1)(Bn+1) = 1/ (X2+1) = -1/( n2+2n) = -0.5*(1/n - 1/(n+2))
转化为减法,将n的数值代入,进行销项,即可得结果
由二次方程根与系数的关系(也就是韦达定理)可得
An+Bn=-(2n+1) ,An*Bn=n^2 ,
因此 1/[(An+1)(Bn+1)]=1/(An*Bn+An+Bn+1)=1/(n^2-2n-1+1)=1/[n(n-2)]=1/2*[1/(n-2)-1/n] ,
所以,原式=1/(1*3)+1/(2*4)+.....+1/(18*20)
=1/2*(1-1/3...
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由二次方程根与系数的关系(也就是韦达定理)可得
An+Bn=-(2n+1) ,An*Bn=n^2 ,
因此 1/[(An+1)(Bn+1)]=1/(An*Bn+An+Bn+1)=1/(n^2-2n-1+1)=1/[n(n-2)]=1/2*[1/(n-2)-1/n] ,
所以,原式=1/(1*3)+1/(2*4)+.....+1/(18*20)
=1/2*(1-1/3+1/2-1/4+1/3-1/5+.......+1/18-1/20)
=1/2*(1+1/2-1/19-1/20)
=531/760 。
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