數列~qqqqqqqq

選取等差數列 {An} = A1, A2,... A(2n+1), 共有 2n+1項

然後, A1 + A3 + ... + A(2n+1) = 319 ---------- (1)

和 A2 + A4 + ... + A(2n) = 290 ---------- (2)

(1) - (2) : A1 + (A3 - A2) + (A5 - A4) + ... + [A(2n+1) - A(2n)] = 29

___________________________________________A1 + nd = 29
(在這裡, d 是 {An} 的等差)

{An} 的一般項是 T(n) = A1 + (n - 1)d, 而 n = 1, 2,... 2n+1

第 n+1 項 : T(n+1) = A1 + [(n+1) - 1]d
________________= A1 + nd
________________= 29