Question

高中数学求解，谢谢

Answer 1

(1).a3-a1=2d=-4,d=-2,an=a1+(n-1)d=1-2(n-1)=3-2n,Sk=(a1+ak)*k/2=(1+3-2k)k/2=(2-k)k=-35
k^2-2k-35=0,(k-7)(k+5)=0,k=7(k=-5舍去)
(2).an=a1+(n-1)d=1-2(n-1)=3-2n.Sk+2-Sk=a(k+2)+a(k+1)=3-2(k+2)+3-2(k+1)=-4k=24,k=-6
因为k为正整数，k不存在。
a6-a3=3d=7-4=3,d=1,由a3=a1+2d,得4=a1+2*1,a1=2,an=a1+(n-1)d=2+(n-1)*1=n+1
1/[a(3n+1)a(3n+4)]=1/[(3n+2)(3n+5)]=[1/(3n+2)-1/(3n+5)]/3;
Tn={(1/5-1/8)+(1/8-1/11)+(1/11-1/14)+...+[1/(3n+2)-1/(3n+5)]}/3
=[1/5-1/(3n+5)]/3
=1/15-1/(9n+15)