记方程(1)x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之和为a3 求a2006
问题描述:
记方程(1)x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之和为a3 求a2006
记方程(1)x^2-3x+2=0的两根之和为a1
(2) x^2+7x+12=0两根之和为a2
(3)x^2-11x+30=0两根之和为a3
求a2006的值
求a1+a2+a3……+a2008
答
由韦达定理 x1+x2=-b/a∴ a1=3,a2=-7,a3=11有 |a2|-|a1|=|a3|-|a2|=4|an|=4n-1an=(4n-1)*(-1)^(n+1),n∈N*∴ a2006=(4*2006-1)(-1)^(2006+1)= -8023S2008=(a1+a2)+(a3+a4)+……+(a2007+a2008)=(-4)*2008/2= -4016...