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

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

an=3n-50
a1=-47
a2=-44
a3=-41
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数列｛an｝为等差,公差d=3
an=-47+3(n-1)=3n-50<=0
n=16时,a16=-2<0,
n=17,a17<0
Sn最小值=s16=(-47-2)*16/2
=-392