一題級數和問題

第一項是1 第二項是5 第三項是13

T(1) = 1

T(2) = 1 + 4 = 5

T(3) = 5 + 8 = 13

T(4) = 13 + 12 = 25

因此T(n) - T(n - 1) = 4(n - 1)

T(n) = T(n - 1) + 4(n - 1)

= T(n - 2) + 4(n - 2) + 4(n - 1)

= ...

= T(1) + 4 + ... + 4(n - 1)

= 1 + 2n(n - 1)

= 2n^2 - 2n + 1

∑ T(n)

= ∑ 2n^2 - 2n + 1

= n(n + 1)(2n + 1)/3 - n(n + 1) + n

前20項的和

= 20(21)(41)/3 - 20(21) + 20

= 5340

T1 = 1 = 0^2 + 1^2
T2 = 5 = 1^2 + 2^2
T3 = 13 = 2^2 + 3^2
. . .
所以 T20 = 19^2 + 20^2
因為 ∑ n^2 = n(n + 1)(2n + 1)/6, 所以前20項的和是 :
19*20*39/6 + 20*21*41/6
= 2470 + 2870
= 5340