有数列{an},a1=5/6,若二次方程a下标(n-1)x^2减a下标(n)x+1=0有根x,y且满足3x-xy+3y=1
问题描述:
有数列{an},a1=5/6,若二次方程a下标(n-1)x^2减a下标(n)x+1=0有根x,y且满足3x-xy+3y=1
1.求证{an-1/2}为等比数列 2.通向公式an 3.前n项和Sn
答
a(n-1)x²-an x+1=0两根分别为x、y.由韦达定理得x+y=an/a(n-1)xy=1/a(n-1)3x-xy+3y=13(x+y)-xy=13an/a(n-1)-1/a(n-1)=1整理,得3an=a(n-1)+13an-3/2=a(n-1)-1/2(an- 1/2)/[a(n-1)-1/2]=1/3,为定值.a1-1/2=5/6-1/2...