Question

已知数列an的通项公式为an=3n-50,求其前n项和Sn的最小值

Answer 1

a1=3-50=-47
an为d=3的等差数列
sn=n(a1+an)/2=n(-47+3n-50)/2=(3n^2-97n)/2
将上式看作连续函数求导
(6n-97)/2=0
n=97/6
此函数二阶导数为3>0，因此在n=97/6为最小值
因此sn在n=16或n=17处取得最小值
当n=16时，sn=-392
当n=17时，sn=-391
所以最小值为-392

Answer 2

a(n) = 3n-50, n = 1,2,...
S(n) = a(1) + a(2) + ... + a(n)
= 3*1 - 50 + 3*2 - 50 + ... + 3*n - 50
= 3[1 + 2 + ... + n] - 50n
= 3n(n+1)/2 - 50n
= 3n^2/2 - 97n/2
= 3/2[n^2 - 97n/3 + (97/6)^2] - 3/2(97/6)^2
= 3/2[n - 97/6]^2 - 97^2/24.

96/6 = 16 < 97/6 < 102/6 = 17

S(16) = 3/2[1/6]^2 - 97^2/24,
S(17) = 3/2[5/6]^2 - 97^2/24 > S(16)

所以，S(n)的最小值=S(16) = 3/2[1/6]^2 - 97^2/24 = 1/24 - 97^2/24 = -96*98/24 = -392

Answer 3

an=3n-50=3(n-1)-47
首项a1=-47，公差d=3
前n项和Sn=na1+n(n-1)d/2=3/2n平方-97/2n
当n=-(197/2)/(3/2)=97/3时，sn最大
但是n必为整数，n=32时，sn=-16，
n=33时，sn=33
因此，n=33时，sn最大，为33

Answer 4

an=3n-50,
a1=-47
d=3
a17=1>0
a16=-2<0
n=16,Sn取最小值:
s16=16*(-47)+16*15*3/2
=-392