等差数列a1为10a2为整数sn小于等于s4求an通项公式,设bn等于anan+一分之一,则bn前n项和?

Question

等差数列a1为10a2为整数sn小于等于s4求an通项公式,设bn等于anan+一分之一,则bn
前n项和?

Answer 1

设公差为d
d=a2-a1=a2-10
a2为整数,d为整数
an=a1+(n-1)d=10+(n-1)d
Sn≤S4,即当n=4时,Sn取得最大值
a4≥0 a5≤0
10+(4-1)d≥0 3d≥-10
d≥-10/3
10+(5-1)d≤0 4d≤-10
d≤-5/2
-10/3≤d≤-5/2
d为整数,d=-3
an=a1+(n-1)d=10+(-3)(n-1)=-3n+13
数列{an}的通项公式为an=-3n+13
bn=1/[ana(n+1)]
=1/[(-3n+13)(-3n+10)]
=1/[(3n-13)(3n-10)]
=(1/3)[1/(3n-10)-1/(3n-13)]
Tn=b1+b2+...+bn
=(1/3)[1/(3×2-13)-1/(3×1-13)+1/(3×3-13)-1/(3×2-13)+...+1/[3(n+1)-13]-1/(3n-13) ]
=(1/3)[1/(3(n+1)-13)-1/(3×1-13)]
=(1/3)[1/(3n-10)+1/10]
=1/(9n-30) +1/30