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

Question

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

Answer 1

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；①求a2006的值；②求a1+a2+a3……+a2008
①a₁=3=1+2,a₂=-7=-3-4,a₃=11=5+6,.,（注意两根之积：1×2=2；(-3)×(-4)=12；5×6=30,.）
a‹n›=(2n-1)×(-1)^(n+1)+2n×(-1)^(n+1)=(4n-1)×(-1)^(n+1),(n=1,2,3,.,n)
(奇数项为正,偶数项为负.)
故a‹2006›=-(4×2006-1)=-8023
②a‹2n-1›+a‹2n›=[4(2n-1)-1]-(4×2n-1)=-5+1=-4
故S‹2008›=-4×1004=-4016