Question

已知数列an的前n项和Sn=-1/2n²+kn,且Sn的最大值为8（1）确定常数k，求an

Answer 1

s(n)=(-1/2)n^2 + kn = (-1/2)[n^2 - 2kn + k^2 - k^2] = (-1/2)(n-k)^2 + (1/2)k^2
= (1/2)[k^2 - (n-k)^2]
<= (1/2)k^2,
s(n)的最大值为(1/2)k^2,
所以，8 = (1/2)k^2, 16 = k^2, k=4或k=-4.

s(n)=(-1/2)n^2 + kn,
a(1)=s(1)=-1/2 + k,
s(n+1)=(-1/2)(n+1)^2 + k(n+1),
a(n+1) = (-1/2)(2n+1) + k = -n + k-1/2,
a(n) = k-1/2 -(n-1) = k - 1/2 - n + 1 = k + 1/2 - n.
k=4时，a(n) = 9/2 - n,
k=-4时，a(n)=-7/2 - n.

Answer 2

Sn=-1/2n²+kn
=-1/2(n²-2kn+k²)+k²/2
=-1/2(n-k)²+k²/2

∵Sn的最大值为8

∴n=k时，Sn取得取得最大值 8
k²/2=8,k²=16,k=4
(-4要舍去的n是正整数不能去-4的)

∴Sn=-1/2n²+4n

a1=S1=7/2
n≥2时，
an=Sn-S(n-1)
=-n+9/2
n=1时上式仍然成立
∴an=-n+9/2

注意本题只有一个结果的！