Question

设正项数列{an}的前n项和是sn,若{an}和{更号sn}都是等差数列,且公差相等,则a1+d=

Answer 1

设公差为d,首项a1

Sn=na1+n(n-1)d/2

√Sn=√(S1)+(n-1)d=√(a1)+(n-1)d 平方

Sn=a1+2√(a1)*(n-1)d+(n-1)^2d^2

na1+n(n-1)d/2=a1+2√(a1)*(n-1)d+(n-1)^2d^2

(n-1)a1+n(n-1)d/2=2√(a1)*(n-1)d+(n-1)^2d^2

a1+nd/2=2√(a1)d+(n-1)d^2

nd/2-nd^2=2√(a1)d-d^2-a1

n(d/2-d^2)=2√(a1)d-d^2-a1

因为对任意n∈N+都成立,所以

d/2-d^2=0

d=0或d=1/2

(1) d=0时 2√(a1)d-d^2-a1=0 a1=0 正项数列 舍

(2) d=1/2 2√(a1)d-d^2-a1=0 a1=1/4