11.解: 由韦达定理,an+a(n+1)=-3n,an*a(n+1)=bn,an+a(n+1)=-3n, a(n-1)+an=-3(n-1),则a(n+1)-a(n-1)=-3,则数列{a1,a3,a5……a(2n-1),a(2n+1)}是以a1为首项,公差为-3的等差数列,数列{a2,a4,a6……a(2n),a(2n+2)}是以a2为首项,公差为-3的等差数列。a(2n-1)=a1+(n-1)(-3); a(2n)=a2+(n-1)(-3),由a10=-17,则a10=a2+(5-1)(-3)=-17,a2=-5又a1+a2=-3,a1=2,b51=a51*a52=[a1+(26-1)(-3)][a2+(26-...
|