Question

等差数列........

已知等差数列an a1大于零，s5等于s13求sn最大时n为多少

Answer 1

Sn = na1 + n(n-1)d/2
S5 = 5a1 + 10d
S13= 13a1+ 78d
S5 = S13
=> 5a1 + 10d = 13a1+ 78d
=> 8a1 + 68d = 0
=> a1 = -17d/2
又a1>0 , ==> d<0
Sn = na1 + n(n-1)d/2
= 1/2*[n(n-1)d - 17nd]
= 1/2*d*[n(n-1) - 17n]
= 1/2*d*[n^2 - 18n]
= 1/2*d*[n^2 - 18n + 81 - 81]
= 1/2*d*[(n - 9)^2 - 81]
= [-1/2*d]*[81 - (n - 9)^2 ]
因d<0, 故[-1/2*d]>0
==> 要Sn最大，也即[81 - (n - 9)^2 ]最大
==> 当n-9=0时，即n=9时，Sn最大
==> 当n = 9时，Sn最大，Sn = -81/2*d

Answer 2

那个麻烦你写的清楚一点可以吗？
我看不太懂~~

Answer 3