Question

等差数列{an}的公差是正数,前n项的和sn,且a3a7=-15,a4+a6=-2,《1》求数列{an}的通项公式及前n

Answer 1

①∵a3a7=(a5-2d)(a5+2d)=-15

∴a5²-4d²=-15

∵a4+a6=2a5=-2

a5=-1

∴1-4d²=-15

d=2

∴an=a5+d(n-5)

=2n-11

②bn=1/n（an+13）

=1/n(2n+2)

=0.5×1/n(n+1)

=0.5×［1/n -1/（n+1）］(裂项法)

∴Tn=0.5×［1-1/2+1/2-1/3+1/3-1/4+…+1/n -1/(n-1)］

=0.5×［1-1/(n-1)］

=1/2 -1/2(n-1)

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