UC知道两道高一数列问题，急！！
来源：百度知道 编辑：UC知道 时间：2024/06/20 09:35:36
(2)等差数列｛An｝的前n项和为Sn，且S9=18,Sn=240,A(n-4)=30,n>9且n属于正整数，求n的值.
A(n) = a + (n-1)d, n = 1,2,...
(1)
10n - n^2 = S(n) = na + n(n-1)d/2 = n(a - d/2) + n^2d/2.
d/2 = -1, d = -2,
a = d/2 + 10 = 9.
A(n) = 9 - 2(n-1) = 11 - 2n
n <= 5, A(n) > 0, B(n) = A(n) = 11 - 2n, n = 1,2,3,4,5.
n > 5, A(n) < 0, B(n) = -A(n) = 2n - 11 = 2(n-5) - 1, n >= 6.
n <= 5, B(1) + B(2) + ... + B(n) = 11n - n(n+1) = n(10-n).
n > 5, B(1) + B(2) + ... + B(n) = 11*5 - 5*(5+1) + 2*1 - 1 + 2*2 - 1 + ... + 2*(n-5) - 1
= 25 + 2[1 + 2 + ... + n-5] - (n-5)
= 25 + (n-5)(n-4) - n + 5
= 25 + (n-5)^2
(2)
S(n) = na + n(n-1)d/2,
18 = S(9) = 9a + 9*8*d/2 = 9a + 36d,2 = a + 4d. a = 2 - 4d
240 = S(n) = na + n(n-1)d/2 = n[2-4d] + n(n-1)d/2,
30 = A(n-4) = a + (n-5)d = 2 - 4d + (n-5)d = 2 + (n-9)d,
n 不等于9.
d = 28/(n-9)
240 = n[2 - 4*28/(n-9)] + n(n-1)*14/(n-9),
240(n-9) = n[2n - 18 - 4*28] + n(n-1)*14,
0 = 16n^2 -