general term~10,6,3,1

T(1) = 10
T(2) = 10 - 4
T(3) = 10 - (4 + 3)
T(4) = 10 - (4 + 3 + 2)
..............................T(n) = 10 - (4 + 3 + 2 + 1 + 0 + -1 + -2 + ...)括弧內有 n-1 項。等差數列和
4 + 3 + 2 + 1 + 0 + -1 + -2 + ...
首項 = 4
公差 d = - 1
項數 = n - 1故尾項 = 4 + (n-2)(- 1) = 6 - n
4 + 3 + 2 + 1 + 0 + -1 + -2 + ... (n-1 項)
= (首項 + 尾項) x 項數 / 2
= (4 + 6-n)(n - 1) / 2
= (10 - n)(n - 1) / 2即
T(n) = 10 - (10 - n)(n - 1) / 2= [20 + (n - 10)(n - 1)] / 2= (n² - 11n + 30) / 2= (n - 5) (n - 6) / 2