题目

已知sn为数列{an}前n项和,a1=a为正整数,sn=ka(n+1),0<绝对值k<1
对于每一个正整数m,若将数列中的三项a(m+1),a(m+2),a(m+3)按从小到大的顺序调整后,均可构成等差数列,且记公差为dm,求k的值和dm的表达式（用含m,a的式子表达）

提问时间：2021-04-01

答案

s(n) = ka(n+1) = k[s(n+1)-s(n)],ks(n+1)= (k+1)s(n),
0< |k| < 1.k不为0.
s(n+1) = [(k+1)/k]s(n).
{s(n)}是首项为s(1)=a(1)=a,公比为(k+1)/k的等比数列.
s(n) = a*[(k+1)/k]^(n-1).
a(n+1) = s(n)/k = (a/k)[(k+1)/k]^(n-1) = [a/(k+1)][(k+1)/k]^n,
a(1)=a,
n>=2时,a(n) = a/(k+1)[(k+1)/k]^(n-1),
a(m+2) = a/(k+1)[(k+1)/k]^(m+1)
a(m+1) = a/(k+1)[(k+1)/k]^m = [k/(k+1)]a(m+2)
a(m+3) = a/(k+1)[(k+1)/k]^(m+2) = [(k+1)/k]a(m+2)
k不为0,-1 1.
a(m+2)＝a/(k+1)[(k+1)/k]^(m+1) >0,
a(m+1) = [k/(k+1)]a(m+2) = a(m+2) - [1/(k+1)]a(m+2) < a(m+2).
a(m+3) = [(k+1)/k]a(m+2) = a(m+2) + (1/k)a(m+2) > a(m+2).
a(m+1) < a(m+2) < a(m+3).
2a(m+2) = a(m+1) + a(m+3) = [k/(k+1)]a(m+2) + [(k+1)/k]a(m+2).
2 = k/(k+1) + (k+1)/k = 1 - 1/(k+1) + 1 + 1/k,k+1=k,无解.
因此,只能-1 0 > a(m+3) > a(m+1).
d(m) = a(m+2)-a(m+3) = (9a/2)(-1/2)^(m+1).
综合,有,
k=-2/3.
d(2m-1) = (9a/2)(-1/2)^(2m) = (9a/2)(1/4)^m.
d(2m) = (-9a/2)(-1/2)^(2m+1) = (9a/4)(1/4)^m.