Question

设数列（an),a1=5/6,若以a1,a2,.,an为系数的二次方程：a(n-1)X2-anX+1=0,都有根A、B满

Answer 1

(1)因为a(n-1)X2-anX+1=0的根为A、B由韦达定理有A+B=-(-an)/a(n-1)=an/a(n-1)AB=1/a(n-1)又因为3A-AB+3B=1即3(an/a(n-1)) - 1/a(n-1)=1两边乘a(n-1),得3an - 1= a(n-1)设3(an + k)=a(n-1) + k展开得3an + 3k=a(n-1) + k,对照上式得2k=-1,即k=-1/2所以有3(an - 1/2)=a(n-1) - 1/2所以(an - 1/2) / (a(n-1) - 1/2)=1/3a1- 1/2=5/6 - 1/2=1/3所以{an-1/2}是以1/3为首项,1/3为公比的等比数列(2)由(1)得an - 1/2 = (1/3)*(1/3)^(n-1)=(1/3)^n所以得an=(1/3)^n + 1/2(3)Sn=a1+a2+a3+...+an=(1/3)^1 + 1/2 + (1/3)^2 + 1/2 + (1/3)^3 + 1/2 + ...+ (1/3)^n + 1/2=[(1/3)^1 + (1/3)^2 + (1/3)^3 + ...+ (1/3)^n]+(1/2)*n=(1/3)*[1 - (1/3)^n]/(1 - 1/3) + n/2=[1 - (1/3)^n]/2 + n/2=[n+1 - (1/3)^n]/2