a(n) + n = -( a(n-1) + n - 1 ) - 2; 怀疑题目出错了 -1是否为+1?
1.假如an表达式中最后一个数字是1,则
b(n) = a(n) + n
b(n) = -b(n-1)
故b(n)是公比为1的等比数列,a(n) = (-1)^(n+1) * 4 + n
2.
(1) n = 2k,k为N 时,sn = n * (n + 1)/2
(2) n = 2k - 1时,sn = 4 + n * (n + 1)/2
a(n) + n = -( a(n-1) + n - 1 ) - 2; 怀疑题目出错了 -1是否为+1?
1.假如an表达式中最后一个数字是1,则
b(n) = a(n) + n
b(n) = -b(n-1)
故b(n)是公比为1的等比数列,a(n) = (-1)^(n+1) * 4 + n
2.
(1) n = 2k,k为N 时,sn = n * (n + 1)/2
(2) n = 2k - 1时,sn = 4 + n * (n + 1)/2