Question

高一数列求

已知an前n项和sn=1/4n^2+2/3n,求an
数列an中，an=2n-14,当n=多少时，sn取到最小值

Answer 1

解(1)
当n=1时，a1=s1=1/4+2/3=3/12+8/12=11/12
当n≥2时，an=S(n+1)-Sn
=1/4n^2+2/3n-[1/4(n-1)^2+2/3(n-1)]
=11/12+(n-1)×1/2=n/2+5/12
故an=n/2+5/12
(2)令an=2n-14≤0
解得n≤7，
而又由a7=0
知当n=6或n=7时，Sn有最小值S6=S7=a1+a2+...+a6
=6（a1+a6）/2
=6*(-12-2)/2
=-42.

Answer 2

(1)
Sn=(1/4)n^2+(2/3)n

n=1, a1=11/12
for n>=2
an = Sn - S(n-1)
=(1/4)(2n-1) + 2/3
= (1/2)n + 5/12
(2)
an >0
2n -14 >0
n > 7
ie
a1,a2,...,a6<0
min Sn = S6
= (a1+a6)3
= (-12-2)3
= -42