MQ30 --- Series

∑(a = 0 to n) ∑(b = 0 to a) b
= ∑(a = 0 to n) (0+a)(a+1)/2
= (1/2) ∑(a = 0 to n) a(a+1)
= (1/2) ∑(a = 0 to n) a(a+1)(a+2 - (a-1))/3
= (1/2) ∑(a = 0 to n) a(a+1)(a+2 - (a-1))/3
= (1/6) ∑(a = 0 to n) [a(a+1)(a+2) - (a-1)a(a+1)]
= (1/6) [n(n+1)(n+2) - (0-1)0(0+1)]
= n (n+1) (n+2) / 6

2012-07-11 19:50:14 補充：
Why ∑(a = 0 to n) [a(a+1)(a+2) - (a-1)a(a+1)]
= [n(n+1)(n+2) - (0-1)0(0+1)] ?

Because
∑(a = 0 to n) [a(a+1)(a+2) - (a-1)a(a+1)]
=
∑(a = 0 to n) a(a+1)(a+2) - ∑(a = 0 to n)(a-1)a(a+1)
=
∑(a = 0 to n) a(a+1)(a+2) - ∑(a = -1 to n-1) a(a+1)(a+2)
=

2012-07-11 19:50:19 補充：
∑(a = 0 to n) a(a+1)(a+2) - [∑(a = -1) a(a+1)(a+2) + ∑(a = 0 to n-1) a(a+1)(a+2)]
=
[∑(a = 0 to n) a(a+1)(a+2) - ∑(a = 0 to n-1) a(a+1)(a+2)] - ∑(a = -1) a(a+1)(a+2)
=
∑(a = n) a(a+1)(a+2) - ∑(a = 0) (a-1)(a)(a+1)
=
n(n+1)(n+2) - (0-1)0(0+1)